Answer
$$ y'(x) =\frac{\sin y+y \sin x}{\cos x-x \cos y}$$
Work Step by Step
Given $$x\sin y - y\cos x=0$$
Differentiate with respect to $x$
\begin{align*}
\frac{d}{dx} \left[x\sin y - y\cos x\right]&=0\\
x\cos yy'(x) +\sin y + y\sin x - y'(x)\cos x &=0 \\
\left [x\cos y -\cos x \right] y'(x)& =-\sin y - y\sin x
\end{align*}
Then
$$ y'(x) =\frac{\sin y+y \sin x}{\cos x-x \cos y}$$
