Given, $f(x)=\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x$ and $I=\int_0^{\pi / 4} \frac{\sin ^2 2 x}{f(x)} d x$
$=\int_0^{\pi / 4} \frac{4 \sin ^2 x \cos ^2 x}{\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x} d x$
$=\int_0^{\pi / 4} \frac{4 \tan ^2 x \sec ^2 x}{\tan ^6 x+2 \tan ^3 x+1} d x$
Now, put $\tan ^3 x=t$, so at $x=0, t=0$ and at $x=\frac{\pi}{4}$
$t=1$ and $3 \tan ^2 x \sec ^2 x d x=d t$
So, $I=\frac{4}{3} \int_0^1 \frac{d t}{t^2+2 t+1}=\frac{4}{3} \int_0^1 \frac{d t}{(t+1)^2}$
$=\left.\frac{4}{3} \frac{-1}{(t+1)}\right|_0 ^1=\frac{-4}{3}\left[\frac{1}{2}-1\right]=\frac{2}{3}$