Answer
$\tan 3x=\dfrac{3\tan x -\tan^3 x}{1-3\tan^2 x}$
Work Step by Step
Consider LHS: $\tan 3x=\tan (2x+x)$
or, $\dfrac{\tan 2 x+\tan x}{1-\tan 2x \tan x}=\dfrac{\dfrac{2\tan x}{1-\tan^2 x}+\tan x}{1-(\dfrac{2\tan x}{1-\tan^2 x}) \tan x}$
or, $\dfrac{\dfrac{2\tan x}{1-\tan^2 x}+\tan x}{1-(\dfrac{2\tan x}{1-\tan^2 x}) \tan x}=\dfrac{2\tan x+\tan x(1-\tan^2 x)}{(1-\tan^2 x)-2\tan^2 x}$
or, $\dfrac{2\tan x+\tan x-\tan^3 x}{1-3\tan^2 x}=\dfrac{3\tan x -\tan^3 x}{1-3\tan^2 x}$
Thus, $\tan 3x=\dfrac{3\tan x -\tan^3 x}{1-3\tan^2 x}$
Hence, the result has been proved.