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.