Use $\tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)=2\tan^{-1}x$ (valid for $|x|<1$).

So $\tan^{-1}y = \tan^{-1}x + 2\tan^{-1}x = 3\tan^{-1}x$.

Using the triple angle formula: $\tan(3\theta)=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$:

$y = \tan(3\tan^{-1}x) = \dfrac{3x-x^3}{1-3x^2}$