$\frac{d}{dx}\left\{{\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)\right\}=$

$\frac{d}{dx}\left\{{\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)\right\}$

Let $x=\frac{1}{2}\tan \theta$

$=\frac{d}{dx}\left\{{\tan }^{-1}\left(\frac{\frac{6}{2}\tan \theta -8.\frac{1}{8}{\tan }^3\theta }{1-12\times \frac{1}{4}{\tan }^2\theta }\right)\right\}$

$=\frac{d}{dx}\left\{{\tan }^{-1}\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)\right\}$

$=\frac{d}{dx}{\tan }^{-1}\left(\tan 3\theta \right)$

$\Rightarrow \frac{d}{dx}3\theta =3\frac{d}{dx}{\tan }^{-1}\left(2x\right)$

$\Rightarrow \frac{3}{1+4x^2}\times 2=\frac{6}{1+4x^2}$