Answer
$$ y'(x) = \frac{3 (x+y)^{2}-2xy\sec^2\left(x^{2} y\right) }{\left [x^{2}\sec^2 \left(x^{2} y\right) -3(x+y)^2 \right] } $$
Work Step by Step
Given $$\tan \left(x^{2} y\right)=(x+y)^{3} $$
Differentiate with respect to $x$
\begin{align*}
\frac{d}{dx} \tan \left(x^{2} y\right)&=\frac{d}{dx} (x+y)^{3} \\
\sec^2 \left(x^{2} y\right)[x^{2} y'(x) +2x y ]&= 3 (x+y)^{2}( 1+y'(x) ) \\
\left [x^{2}\sec^2 \left(x^{2} y\right) -3(x+y)^2 \right] y'(x)& =3 (x+y)^{2}-2xy\sec^2 \left(x^{2} y\right)
\end{align*}
Then
$$ y'(x) = \frac{3 (x+y)^{2}-2xy\sec^2\left(x^{2} y\right) }{\left [x^{2}\sec^2 \left(x^{2} y\right) -3(x+y)^2 \right] } $$
