We assume that
\begin{align*}
dX_t &= X_t(rdt + \sigma_x dW^x_t)\\
dY_t &= Y_t\Big[rdt + \sigma_y\Big(\rho dW^x_t + \sqrt{1-\rho^2} dW^y_t \Big)\Big],
\end{align*}
where $\rho$ is the correlation, which we assume to be less than 1, and $W^x$ and $W^y$ are two independent standard Brownian motions. Then
\begin{align*}
d(X_t/Y_t) &= \frac{1}{Y_t}dX_t - \frac{X_t}{Y_t^2}dY_t +\frac{X_t}{Y_t^3}d\langle Y, \, Y\rangle_t - \frac{1}{Y_t^2}d\langle X, \, Y\rangle_t\\
&=\frac{X_t}{Y_t}\Big[\sigma_x dW^x_t - \sigma_y\Big(\rho dW^x_t + \sqrt{1-\rho^2} dW^y_t \Big) +\big(\sigma_y^2-\rho\sigma_x\sigma_y\big)dt\Big]\\
&=\frac{X_t}{Y_t}\Big[ \big(\sigma_y^2-\rho\sigma_x\sigma_y\big)dt +\sigma dW_t\Big],
\end{align*}
where
$$\sigma = \sqrt{\sigma_x^2+\sigma^2_y-2\rho\sigma_x\sigma_y},$$ and $W$ is standard Brownian motion.
The option payoff $$(X_T/Y_T-K)^+$$ at maturity $T$ has the value given by
\begin{align*}
C &\equiv e^{-rT}E\big((X_T/Y_T-K)^+ \big)\\
&=e^{-rT}\left(e^{(\sigma_y^2-\rho\sigma_x\sigma_y)T} \frac{X_0}{Y_0}N(d_1)-KN(d_2)\right)\\
&= e^{-rT +(\sigma_y^2-\rho\sigma_x\sigma_y)T}\left(\frac{X_0}{Y_0}N(d_1)-e^{-(\sigma_y^2-\rho\sigma_x\sigma_y)T} KN(d_2)\right),
\end{align*}
where
\begin{align*}
d_1 &=\frac{\ln (X_0/Y_0) + (\sigma_y^2-\rho\sigma_x\sigma_y)T + \frac{1}{2}\sigma^2 T }{\sigma \sqrt{T}},
\end{align*}
and
$$d_2 = d_1 - \sigma \sqrt{T}.$$
Consequently, the delta hedge ratios with respect to $X_0$ and $Y_0$ can be computed correspondingly. Specifically,
\begin{align*}
\frac{\partial C}{\partial X_0} &= \frac{1}{Y_0}\frac{\partial C}{\partial (X_0/Y_0)}\\
&=\frac{1}{Y_0}e^{-rT +(\sigma_y^2-\rho\sigma_x\sigma_y)T}N(d_1),
\end{align*}
and
\begin{align*}
\frac{\partial C}{\partial Y_0} &= -\frac{X_0}{Y_0^2}\frac{\partial C}{\partial (X_0/Y_0)}\\
&=-\frac{X_0}{Y_0^2}e^{-rT +(\sigma_y^2-\rho\sigma_x\sigma_y)T}N(d_1).
\end{align*}