Differentiate $\tan^3(x^2)$ 
Differentiate $\tan^3(x^2)$

I first applied the chain rule and made $u=x^2$ and $g=\tan^3u$. I then calculated the derivative of $u$, which is $$u'=2x$$ and the derivative of $g$, which is 
$$g'=3\tan^2u$$
I then applied the chain rule and multiplied them together, which gave me 
$$f'(x)=2x3\tan^2(x^2)$$
Is this correct? If not, any hints as to how to get the correct answer?

Your answer is not correct because $g' \neq 3\tan^2u$. For getting $g'$ you need to apply the chain rule once more.
$$g(u) = f(h(u)) = f(\tan u) = \tan^3u$$
where $f(x) = x^3$ and $h(u) = \tan u$.
So $g'(u) = f'(h(u))*h'(u) = 3(h(u))^2*\sec u = 3\tan^2u * \sec u$
