The question specified

Find $\frac{dy}{dx}$ of $\tan^{-1}(\frac{3x - x^3}{1 - 3x^2})$

However, we were asked to try the sum without using the trig identities of $\tan(3x)$. I used chain rule and the general formula of $\frac{d}{dx}\tan^{-1}(x) = \frac{1}{1 + x^2}$ and after few steps I get the expression $\frac{6x^2 + 3x^4 + 3}{3x^4 + 3x^2 + x^6 + 1}$ and this is where I cannot further simplify. Can someone please point out how to proceed or is the only way to do this sum using the trig identities ?