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Assume there are two stocks $S_1$ with price $p_1(t)$ and $S_2$ with price $p_2(t)$ where $t$ indicates time. Assume, there is a hypothetical derivative $D$, which is such that, price of $D$ at a time $t$ is given by $p_1(t)/p_2(t)$.
Is it possible to find a dynamic hedging strategy using stocks and bonds to construct this derivative?
If yes, then how?
Let assume that $S_1(t)$ and $S_2(t)$ are the price processes of your 2 stocks and that they follow a Geometric Brownian Motion (GBM):
$$\forall \, i \in \{1,2\}, dS_i(t) =\mu_iS_i(t)dt + \sigma_iS_i(t)dW_i(t)$$
We assume both stocks have an instant correlation of $\rho$:
$$dW_1(t)dW_2(t)=\rho dt$$
Let also $V(t)$ be the value or (fair) price of a derivative that depends on prices $S_1(t)$ and $S_2(t)$ at time $t$. We construct a self-financing portfolio made up of $w_0(t)$ derivative contracts, $w_1(t)$ shares of stock $1$ and $w_2(t)$ shares of stock $2$. Its value at $t$, $\Pi(t)$, is given by:
$$\Pi(t) = w_0(t)V(t)+w_1(t)S_1(t)+w_2(t)S_2(t)$$
The portfolio value, being self-financing, evolves according to:
$$ d\Pi(t) = w_0(t)dV(t) + w_1(t)dS_1(t) + w_2(t)dS_2(t) $$
We then have $-$ dropping time:
$$dV = \frac{\partial V}{\partial t}dt + \frac{\partial V}{\partial S_1}dS_1 + \frac{\partial V}{\partial S_2}dS_2 + \frac{1}{2}\frac{\partial^2 V}{\partial S_1^2}dS_1^2 + \frac{1}{2}\frac{\partial^2 V}{\partial S_2^2}dS_2^2 + \frac{\partial^2 V}{\partial S_1 \partial S_2}dS_1dS_2$$
In the above differential equation, multiplied back by weight $w_0(t)$, the random element is:
$$ \frac{\partial V}{\partial S_1}w_0\sigma_1S_1dW_1 + \frac{\partial V}{\partial S_2}w_0\sigma_2S_2dW_2 $$
Now, in $w_1(t)S_1(t)+w_2(t)S_2(t)$, the random element is:
$$ w_1\sigma_1S_1dW_1 + w_2\sigma_2S_2dW_2 $$
We are hedging the derivative $V(t)$, hence our portfolio must be riskless and earn the risk-free rate:
$$ \begin{align} w_0(t) & = \frac{\Pi(t)}{V(t)-S_1(t)\frac{\partial V}{\partial S_1}-S_2(t)\frac{\partial V}{\partial S_2}} \\[12pt] w_1(t) &= -w_0(t)\frac{\partial V}{\partial S_1} \\[12pt] w_2(t) &= -w_0(t)\frac{\partial V}{\partial S_2} \end{align} $$
$$d\Pi(t) = r\Pi(t) dt$$
The final expression of the hedging portfolio is:
$$ \begin{align} w_0(t) & = \frac{B(t)}{V(t)-S_1(t)\frac{\partial V}{\partial S_1}-S_2(t)\frac{\partial V}{\partial S_2}} \\[12pt] w_1(t) &= -w_0(t)\frac{\partial V}{\partial S_1} \\[12pt] w_2(t) &= -w_0(t)\frac{\partial V}{\partial S_2} \end{align} $$
Now, in the particular case of your derivative and interpreting strictly your original question:
" [...] price of [the derivative] $D$ at a time $t$ is given by $\frac{S_1(t)}{S_2(t)}$. "
The price is then given by:
$$ V(t) = \frac{S_1(t)}{S_2(t)}$$
We have $\partial V/\partial S_1 = 1/S_2 = V/S_1$ and $\partial V/\partial S_2 = -S_1/S_2^2 = -V/S_2$, hence the hedging strategy would be:
$$ \begin{align} w_0(t) &= \frac{B(t)}{V(t)} \\[12pt] w_1(t) &= -\frac{B(t)}{V(t)}\frac{\partial V}{\partial S_1} = -w_0(t)\frac{\partial V}{\partial S_1} \\[12pt] w_2(t) &= -\frac{B(t)}{V(t)}\frac{\partial V}{\partial S_2} = -w_0(t)\frac{\partial V}{\partial S_2} \end{align} $$
However, I am not sure that positing the price $V(t)$ is the correct approach: the price of the derivative at $t$ should be derived from its payoff function and the PDE resulting from the risk-free return condition. After a few steps, we would get the following PDE $-$ dropping time:
$$ \begin{align} rV & = \frac{\partial V}{dt} + \frac{\partial V}{\partial S_1}rS_1 + \frac{1}{2}\frac{\partial^2V}{\partial S_1^2}\sigma_1^2S_1^2 + \frac{\partial V}{\partial S_2}rS_2 + \frac{1}{2}\frac{\partial^2V}{\partial S_2^2}\sigma_2^2S_2^2 + \frac{\partial^2V}{\partial S_1 \partial S_2}\sigma_1\sigma_2S_1S_2\rho \\[12pt] & = \frac{\partial V}{dt} + rw_1S_1 + \frac{1}{2}\frac{\partial^2V}{\partial S_1^2}\sigma_1^2S_1^2 + rw_2S_2 + \frac{1}{2}\frac{\partial^2V}{\partial S_2^2}\sigma_2^2S_2^2 + \frac{\partial^2V}{\partial S_1 \partial S_2}\sigma_1\sigma_2S_1S_2\rho \end{align}$$
Solving it would yield the value of $V(t)$. Note that weight $w_0$ does not appear in the PDE above because all three weights $w_0$, $w_1$ and $w_2$ can be written as a function of $w_0$, hence the term $w_0$ ends up being cancelled.
In this particular case, by replacing the derivatives by their specific expression $-$ which we can derive from the fact that $V(t) = S_1(t)/S_2(t)$ $-$ we obtain:
$$ \begin {align} & \: rV = 0 + rV + 0 - rV + \sigma_2^2V - \sigma_1\sigma_2\rho V \\[12pt] \Leftrightarrow & \: r = \sigma_2^2 - \sigma_1\sigma_2\rho \end{align}$$
Hence we would be imposing a constraint on "market" parameters. The correct way to proceed would be to derive an expression for $V(t)$ given the PDE above.
Let assume that your derivative has the following payoff function:
$$ V_T=V(T)=\frac{S_1(T)}{S_2(T)} $$
Switching to martingale pricing tools, we know that:
$$ \forall \, t \in [0,T], V_t = \mathbb{E}^{\mathbb{Q}}\left[e^{-r(T-t)}\frac{S_1(T)}{S_2(T)}|\mathcal{F}_t\right] $$
Where $\mathbb{Q}$ is the risk-neutral measure. Let now define $X_t = X(t) = S_1(t)/S_2(t)$. Applying Ito's lemma $-$ and dropping time:
$$ \begin{align} dX_t = \frac{dS_1}{S_2} - \frac{S_1dS_2}{S_2^2} + \frac{S_1dS_2^2}{S_2^3} - \frac{dS_1dS_2}{S_2^2} \\[12pt] \Leftrightarrow \frac{dX_t}{X_t} = \left(\mu_1-\mu_2-\sigma_1\sigma_2\rho+\sigma_2^2\right)dt +\sigma_1dW_1^{\mathbb{Q}} + \sigma_2dW_2^{\mathbb{Q}} \end{align} $$
Under the risk-neutral measure, $\mu_1=\mu_2=r$. Letting $W_i(t) = W_i^{\mathbb{Q}}(t)$, after a few steps we get:
$$ \begin{align} X_T = X_te^{(\sigma_2^2-\sigma_1^2-4\sigma_1\sigma_2\rho)\frac{T-t}{2}+\sigma_1W_1(T-t)+\sigma_2W_2(T-t)} \end{align} $$
Applying Ito's lemma to $W_1(t)W_2(t)$ and using the martingale property of the Ito integral, we get:
$$ \mathbb{Cov}[W_1(T-t),W_2(T-t)] = \rho (T-t)$$
Now, letting:
$$ \begin{align} & Z(t) = (\sigma_2^2-\sigma_1^2-4\sigma_1\sigma_2\rho)\frac{t}{2}+\sigma_1W_1(t)+\sigma_2W_2(t) \\[6pt] & \mathbb{E}[Z(t)] = (\sigma_2^2-\sigma_1^2-4\sigma_1\sigma_2\rho)\frac{t}{2} \\[6pt] & \mathbb{V}[Z(t)] = \sigma_1^2t + \sigma_2^2t+2\sigma_1\sigma_2\rho t \end{align} $$
We obtain:
$$ \mathbb{E}[X_T|\mathcal{F}_t] = X_te^{\sigma_2^2(T-t)-\sigma_1\sigma_2\rho (T-t)} $$
Hence
$$ \forall \, t \in [0,T], V_t = \frac{S_1(t)}{S_2(t)}e^{(\sigma_2^2-\sigma_1\sigma_2\rho-r)(T-t)} $$
We have came back to the constraint on market parameters derived in section "Your question":
$$ r = \sigma_2^2 - \sigma_1\sigma_2\rho \quad \Rightarrow \quad V_t = \frac{S_1(t)}{S_2(t)}$$
The hedging strategy would then be:
$$ \begin{align} & w_0(t) = \frac{S_2(t)}{S_1(t)}e^{(\sigma_1\sigma_2\rho-\sigma_2^2)(T-t)+rT} \\[12pt] & w_1(t) = -\frac{e^{rt}}{S_1(t)} \\[12pt] & w_2(t) = \frac{e^{rt}}{S_2(t)} \end{align} $$
You can easily check that $w_0(t)V(t)+w_1(t)S_1(t)+w_2(t)S_2(t) = B(t)$.
We finally can check that:
$$ \begin{align} & \frac{\partial V}{dt} = -(\sigma_2^2-\sigma_1\sigma_2\rho-r)V \\[12pt] & \frac{\partial V}{\partial S_1}rS_1 = rV \\[12pt] & \frac{1}{2}\frac{\partial^2V}{\partial S_1^2}\sigma_1^2S_1^2 = 0 \\[12pt] & \frac{\partial V}{\partial S_2}rS_2 = -rV \\[12pt] & \frac{1}{2}\frac{\partial^2V}{\partial S_2^2}\sigma_2^2S_2^2 = \sigma_2^2V \\[12pt] & \frac{\partial^2V}{\partial S_1 \partial S_2}\sigma_1\sigma_2S_1S_2\rho = -\sigma_1\sigma_2\rho V \end{align} $$
Terms cancel and we are left with the following identity equation:
$$ rV = rV $$
Hence $V(t)$ as derived above is a solution to the pricing PDE.
Further relevant references from Wikipedia and user @Gordon on deriving the hedging strategy and the pricing PDE for a derivative $V(t)$:
Black-Scholes equation (Wikipedia)
Black Scholes differential (Stack Exchange Quantitative Finance)
Derivation of BS PDE problem using Delta hedging (Stack Exchange Quantitative Finance)
