Given, $sin\,y=x\,sin \left(a+y\right)$
$\Rightarrow x=\frac{sin\,y}{sin\left(a+y\right)}$
On differentiating w.r. to y, we get
$\frac{dx}{dy}=\frac{d}{dy}\left[\frac{sin\,y}{sin\left(a+y\right)}\right]$
$=\frac{sin\left(a+y\right)cos\,y-sin\,y\,cos\left(a+y\right)}{sin^{2} \left(a+y\right)}$
$=\frac{sin\,a}{sin^{2} \left(a+y\right)} \Rightarrow \frac{dy}{dx}=\frac{sin^{2} \left(a+y\right)}{sin\,a}$