If $y={\mathrm{Tan}}^{-1}\left(\frac{4\sin 2x}{\cos 2x-6{\sin }^{2}x}\right)\text{ then }\frac{dy}{dx}\text{ at }x=0\text{ is }$

$y={\tan }^{-1}\left(\frac{8\sin x\cos x}{{\cos }^{2}x-7{\sin }^{2}x}\right)$

$={\tan }^{-1}\left(\frac{8\tan x}{1-7{\tan }^{2}x}\right)={\tan }^{-1}\left(\frac{7\tan x+\tan x}{1-7\tan x\cdot \tan x}\right)$

$y={\tan }^{-1}\left(7\tan x\right)+{\tan }^{-1}\left(\tan x\right)$

$y=={\tan }^{-1}\left(7\tan x\right)+x$

$\frac{dy}{dx}=\frac{1}{1+{\left(7\tan x\right)}^2}\times 7se{c}^{2}x+1$
$x=0\Rightarrow \frac{dy}{dx}=\frac{1}{1}\times 7+1=8$