The value of $\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x$, is

$-\frac{1}{3\left(1+\tan ^3 x\right)}$

We have,

$I=\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x$

$\Rightarrow I=\int \frac{\tan ^2 x \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x$             [Dividing $N^r$ and $D^r$ by $\cos ^6 x$]

$\Rightarrow I=\frac{1}{3} \int \frac{3 \tan ^2 x \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x$

$\Rightarrow I=\frac{1}{3} \int \frac{1}{\left(1+\tan ^3 x\right)^2} d\left(1+\tan ^3 x\right)$

$=-\frac{1}{3\left(1+\tan ^3 x\right)}+C$