1. Given equation

We have $an^{-1} y=\tan^{-1} x+\tan^{-1}\left(\frac{2x}{1-x^2}\right),$ with $|x|<\frac{1}{\sqrt{3}}.$

We need to find $y$.

2. Recognize the second term

Using the identity $\tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta},$ if we put $\theta=\tan^{-1}x,$ then $\tan 2\theta=\frac{2x}{1-x^2}.$

So, $\tan^{-1}\left(\frac{2x}{1-x^2}\right)=2\tan^{-1}x,$ provided the principal value matches.

Since $|x|<\frac{1}{\sqrt 3},$ let $\theta=\tan^{-1}x$. Then $|\theta|<\tan^{-1}\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{6}.$ Hence $|2\theta|<\frac{\pi}{3}<\frac{\pi}{2},$ so indeed $\tan^{-1}\left(\frac{2x}{1-x^2}\right)=2\theta=2\tan^{-1}x.$

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

3. Take tangent on both sides

Let $\theta=\tan^{-1}x \implies \tan\theta=x.$ Then $\tan^{-1}y=3\theta \implies y=\tan 3\theta.$

Now use the triple-angle formula: $\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.$ Since $\tan\theta=x$, $y=\frac{3x-x^3}{1-3x^2}.$

4. Match with options

Thus a value of $y$ is $\boxed{\frac{3x-x^3}{1-3x^2}}.$

This corresponds to:

• Option C

5. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.