Find $\frac {dy}{dx}$ in the following: $y = tan^{-1} \left(\frac{3x - x^3}{1 - 3x^2}\right), - \; { \frac{1}{\sqrt3} } < x < \;{\frac{1}{\sqrt3} }\\ $

Question

$\begin{array}{1 1} \large \frac{3}{1+x^2} \\ \large \frac{3}{1-x^2} \\ \large \frac{3x-x^3}{1-3x^2} \\ -\large \frac{3x-x^3}{1-3x^2} \end{array}$