Solve this

Here

$y=x \sin (a+y)$

Differentiating it with respect to $x$ using the chain rule and product rule,

$\frac{d y}{d x}=x \frac{d}{d x} \sin (a+y)+\sin (a+y) \frac{d x}{d x}$

$\frac{d y}{d x}=x \cos (a+y) \frac{d y}{d x}+\sin (a+y)$

$(1-x \cos (a+y)) \frac{d y}{d x}=\sin (a+y)$

$\frac{d y}{d x}=\frac{\sin (a+y)}{(1-x \cos (a+y))}$

$\frac{d y}{d x}=\frac{\sin (a+y)}{\left(1 \frac{y}{\sin (a+y)} \cos (a+y)\right)}\left[\right.$ Since, $\left.\frac{y}{\sin (a+y)}=x\right]$

$\frac{d y}{d x}=\frac{\sin (a+y)}{\left(\frac{\sin (a+y)-\cos (a+y) \cdot y}{\sin (a+y)}\right)}$

$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$

Hence Proved.