Answer
$$y' =\frac{y-2x\cos y}{2 \cos 2y -x^2\sin y-x}$$
Work Step by Step
Given
$$ x^2 \cos y + \sin 2y =xy$$
differentiate with respect to $x$ , then
\begin{align*}
x^2( -\sin y)y'+2x\cos y +2 \cos 2y y'&=xy'+y\\
y'( 2 \cos 2y -x^2\sin y-x)&=y-2x\cos y\\
y'&=\frac{y-2x\cos y}{2 \cos 2y -x^2\sin y-x}
\end{align*}