$\begin{aligned} & y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right) \\ & =\tan ^{-1}\left[\frac{4(2 \sin x \cos x)}{\left(\cos ^2 x-\sin ^2 x\right)-6 \sin ^2 x}\right] \\ & =\tan ^{-1}\left(\frac{8 \sin x \cos x}{\cos ^2 x-7 \sin ^2 x}\right)\end{aligned}$
$\begin{aligned} & =\tan ^{-1}\left(\frac{8 \tan x}{1-7 \tan ^2 x}\right) \\ & =\tan ^{-1}\left(\frac{7 \tan x+\tan x}{1-7 \tan x \cdot \tan x}\right) \\ & =\tan ^{-1}(7 \tan x)+\tan ^{-1}(\tan x) \\ \therefore \quad & y=\tan ^{-1}(7 \tan x)+x \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+(7 \tan x)^2} \cdot 7 \sec ^2 x+1\end{aligned}$
$\begin{gathered}=\frac{7 \sec ^2 x}{1+49 \tan ^2 x}+1 \\ \therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}= \\ =\frac{7 \sec ^2 0}{1+49 \tan ^2 0}+1 \\ =\frac{7}{1+0}+1 \\ =8\end{gathered}$