Stochastic Vol Mathematical derivation [closed]

I want to understand the mathematical steps done. Can someone please simplify the derivation of d(pi) from Pi? Thanks in advance.

Let $$f=f(t,s,v)\in C^{1,2,2}(\mathbb{R}_+^3)$$ be a real-valued function (portfolio value) and consider the two-dimensional stochastic process $$(S_t,v_t)$$ with \begin{align*} \mathrm{d}S_t&=(r-q) S_t \mathrm{d}t+\sqrt{v_t} S_t \mathrm{d}W_{1,t}, \\ \mathrm{d}v_t&=\kappa(\theta-v_t) \mathrm{d}t+\xi \sqrt{v_t} \mathrm{d}W_{2,t}, \end{align*} with $$\mathbb{E}[\mathrm{d}W_{1,t}\mathrm{d}W_{2,t}]=\rho\mathrm{d}t$$. Then, denoting partial derivatives by subscripts, we obtain from Ito's Lemma (which byouness mentioned in his comment) \begin{align*} \mathrm{d}f &= \left(f_t + \frac{1}{2}v_tS_t^2f_{ss} + \frac{1}{2}\xi^2v_tf_{vv}+\rho\xi S_tv_tf_{sv}\right) \mathrm{d}t + f_s \mathrm{d}S_t + f_v\mathrm{d}v_t. \end{align*} Following their definition as SDEs, the changes $$\mathrm{d}S_t$$ and $$\mathrm{d}v_t$$ can also be expressed in terms of $$\mathrm{d}W_{1,t}$$ and $$\mathrm{d}W_{2,t}$$ yielding \begin{align*} \mathrm{d}f &= \left(f_t + (r-q)S_tf_s + \kappa(\theta-v_t)f_v+ \frac{1}{2}v_tS_t^2f_{ss} + \frac{1}{2}\xi^2v_tf_{vv}+\rho\xi S_tv_tf_{sv}\right) \mathrm{d}t \\ & \;\;\;\;\; + f_s \sqrt{v_t}S_t \mathrm{d}W_{1,t} + f_v\xi\sqrt{v_t}\mathrm{d}W_{2,t}. \end{align*} The rest is identical to the derivation of the Black Scholes equation, we shall assume that both sources of risks can be eliminated by dynamic hedging forcing $$\mathrm{d}f$$ to be proportional to $$\mathrm{d}t$$. (choose the values you hold in the two assets such that the stochastic terms are zero). Thus, changes in $$f$$ are locally risk-free and hence, $$\mathrm{d}f=rf\mathrm{d}t$$.
We end up with the following linear, second-order, three-dimensional PDE \begin{align*} \frac{\partial f}{\partial t}+\frac{1}{2}v_tS_t^2\frac{\partial f^2}{\partial S_t^2}+\rho\xi v_tS_t\frac{\partial f^2}{\partial S_t\partial v_t}+\frac{1}{2}\xi^2v_t\frac{\partial f^2}{\partial v_t^2}+(r-q)S_t\frac{\partial f}{\partial S_t}+\kappa(\theta-v_t)\frac{\partial f}{\partial v_t}-rf=0. \end{align*}
• @KeSchn a general follow up question. In the screenshot in question, if you make the $dv$ term to zero using $\Delta_1 = \partial V / \partial V_1$ it looks like $\Delta$ should always be zero. Would this be correct? Or is $\frac{\partial V}{\partial V_1} \cdot \frac{\partial V_1}{\partial S}$ not simplifiable – Slade Oct 14 '19 at 21:07