If $\mathrm{x} \mathrm{siny}=3\mathrm{siny}+4 \mathrm{cosy}$ then $\frac{\mathrm{dy}}{\mathrm{dx}}=$

The given equation is
$$\begin{gathered} x\sin y=3\sin y+4\cos y \\ \Rightarrow \left(x-3\right)\sin y=4\cos y \end{gathered}$$
 $$\begin{gathered} \Rightarrow x-3=4coty \\ \Rightarrow x=4coty+3 \\ differentiate w.r.t y \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{dx}{dy}=-4\cos e{c}^{2}y=\frac{-4}{{\sin }^{2}y} \\ \Rightarrow \frac{dy}{dx}=\frac{-{\sin }^{2}y}{4} \end{gathered}$$