If $\sin (y)=x\sin (a+y)$, prove that: If $\sin (y)=x\sin (a+y)$, prove that:
$$\dfrac {dy}{dx}=\dfrac {\sin^2 (a+y)}{\sin (a)}$$
My Attempt:
$$\sin (y)=x\sin (a+y)$$
$$\dfrac {d}{dy} \sin (y)=\dfrac {d}{dy} (x\sin (a+y))$$
$$\cos (y)=x\cdot \cos (a+y)+\sin (a+y)\cdot\dfrac {dx}{dy}$$
$$\cos (y)-x\cdot\cos (a+y)=\sin (a+y)\cdot\dfrac {dx}{dy}$$

By implicit differentiation, you should get
\begin{align}
\cos(y) y' = \sin(a+y) + x\cos(a+y)y' \ \ \implies \ \ y' = \frac{\sin(a+y)}{\cos(y)-x\cos(a+y)}.
\end{align}
However, since
\begin{align}
x= \frac{\sin(y)}{\sin(a+y)} \ \ \implies \ \ y' = \frac{\sin^2(a+y)}{\sin(a+y)\cos(y) - \sin(y)\cos(a+y)} = \frac{\sin^2(a+y)}{\sin(a)}
\end{align}
where the last equality uses the difference angle formula.
