Answer
$3x^4-\dfrac{27}{3}x^{12}+\dfrac{243}{5}x^{20}-\dfrac{2187}{7}x^{28}$
Work Step by Step
Consider the Taylor's Series for $ \tan^{-1} x$ as follows:
$ \tan^{-1} x=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+...$
Plug $(3x^4)$ instead of $x$ in the above series as follows:
Then $ \tan^{-1} (3x^4)=(3x^4)-\dfrac{(3x^4)^3}{3}+\dfrac{(3x^4)^5}{5}-\dfrac{(3x^4)^7}{7}+...$
$ \implies 3x^4-\dfrac{27}{3}x^{12}+\dfrac{243}{5}x^{20}-\dfrac{2187}{7}x^{28}$
