If $f(x)=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)$ then $\frac{d}{d x}(f(x))$ is equal to

$\frac{3}{1+x^2}$ for all $x \in R-\left\{-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\}$

We have,

$f(x)=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)$

$\Rightarrow f(x)= \begin{cases}3 \tan ^{-1} x+\pi & \text { for } x<-\frac{1}{\sqrt{3}} \\ 3 \tan ^{-1} x & \text { for }-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}} \\ 3 \tan ^{-1} x-\pi & \text { for } x>\frac{1}{\sqrt{3}}\end{cases}$

∴  $f'(x)=\frac{3}{1+x^2}$  for all  $x \in R-\left\{-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\}$