Differentiate $\displaystyle tan^32x$ with respect to $\displaystyle x$
$\displaystyle y = \tan^3{2x}$.
Let $\displaystyle u = \tan{2x}$ so that $\displaystyle y = u^3$.
$\displaystyle \frac{du}{dx} = 2\sec^2{2x}$.
$\displaystyle \frac{dy}{du} = 3u^2 = 3\tan^2{2x}$.
Therefore $\displaystyle \frac{dy}{dx} = 6\sec^2{2x}\tan^2{2x}$.
You could use the Pythagorean Identity to clean this up further if you wanted...