$\begin{array}{l}f\left(x\right)=\frac{\tan 2x}{6\cos x+2\sin 2x}isdefinediff\\ \Rightarrow 6\cos x+2\sin 2x\ne 0\\ \Rightarrow 6\cos x+2(2\sin x.\cos x)\ne 0\\ \Rightarrow 2\cos x(3+2\sin x)\ne 0\\ \Rightarrow \cos x\ne 0,3=2\sin x\ne 0\\ \therefore x\ne (2n+1)\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\sin x\ne \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.Whichisnotposible\end{array}$

$\begin{array}{l}and\tan 2x\ne \infty \\ 2x\ne (2n+1)\frac{\pi }{2}\\ x\ne (2n+1)\frac{\pi }{4}\\ \therefore R-\left\{(2n+1)\frac{\pi }{2}\right\}\cup \left\{(2n+1)\frac{\pi }{4}\right\}\end{array}$