$\begin{aligned} & \text { LHS }=\tan x+\tan \left(x+\frac{\pi}{3}\right)+\tan \left(x+\frac{2 \pi}{3}\right) \\ & =\tan x+\frac{\tan x+\sqrt{3}}{1-\sqrt{3} \tan x}+\frac{\tan x-\sqrt{3}}{1+\sqrt{3} \tan x} \\ & =\tan x+\frac{\left[\begin{array}{c}(\tan x+\sqrt{3})(1+\sqrt{3} \tan x) \\ +(\tan x-\sqrt{3})(1-\sqrt{3} \tan x)\end{array}\right]}{1-3 \tan ^2 x} \\ & =\frac{\left[\begin{array}{c}\tan x\left(1-3 \tan ^2 x\right)+\tan x+\sqrt{3} \\ +\sqrt{3} \tan ^2 x+3 \tan x+\tan x-\sqrt{3} \\ -\sqrt{3} \tan ^2 x+3 \tan x\end{array}\right]}{1-3 \tan ^2 x} \\ & \Rightarrow \frac{\tan x\left(1-3 \tan ^2 x\right)+8 \tan x}{1-3 \tan ^2 x}=3 \quad \text { (given) } \\ & \Rightarrow \tan x\left(1-3 \tan ^2 x\right)+8 \tan x \\ & =3\left(1-3 \tan ^2 x\right) \\ & \end{aligned}$
$\begin{aligned} & \Rightarrow \quad \tan x\left(9-3 \tan ^2 x\right)=3\left(1-3 \tan ^2 x\right) \\ & \Rightarrow \quad \tan x\left(3-\tan ^2 x\right)=1-3 \tan ^2 x \\ & \Rightarrow \quad \frac{3 \tan x-\tan ^3 x}{1-3 \tan ^2 x}=1 \\ & \Rightarrow \quad \tan 3 x=1\end{aligned}$