Let $\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$. Then y is:

$\frac{3 x-x^3}{1-3 x^2}$

The correct answer is Option (1) - $\frac{3 x-x^3}{1-3 x^2}$

$\tan^{-1} y=\tan^{-1} x+\tan^{-1}\left(\frac{2 x}{1-x^2}\right)$

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

so $\tan^{-1} y=3\tan^{-1} x$

so $y=\frac{3 x-x^3}{1-3 x^2}$