Find the derivative of the $y=\tan^{-1}(\frac{3x-x^3}{1-3x^3})$

$\frac{1}{1+x}$ $x(1+x^2)$ $-1/x$ $\frac{3}{1+x^2}$

$\frac{3}{1+x^2}$

Put $x=\tan \theta$. Then $y=\tan^{-1}(\tan 3\theta)=3\theta=3\tan^{-1}x$. Differentiating w.r.to x we get $\frac{dy}{dx}=\frac{3}{1+x^2}$