Let $$\tan^{-1} y = \tan^{-1} x + \tan^{-1}\left(\frac{2x}{1 - x^2}\right)$$, where $$|x| \lt \frac{1}{\sqrt{3}}$$. Then a value of $$y$$ is

We are given the relation $$\tan^{-1}y=\tan^{-1}x+\tan^{-1}\!\left(\dfrac{2x}{1-x^{2}}\right)$$ with the condition $$|x|\lt \dfrac{1}{\sqrt{3}}$$ and we have to find an explicit expression for $$y$$.

First, recall the standard sum-of-inverse-tangents formula:

$$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\!\left(\dfrac{A+B}{1-AB}\right),$$

which is valid when the product $$AB\lt 1$$, a condition that will indeed be satisfied here because of the given bound on $$x$$.

We now apply the formula to the two inverse-tangent terms on the right-hand side. Identifying

$$A=x,\qquad B=\dfrac{2x}{1-x^{2}},$$

we can write

$$\tan^{-1}y=\tan^{-1}\!\left(\dfrac{A+B}{\;1-AB\;}\right).$$

So we need to compute both the numerator $$A+B$$ and the denominator $$1-AB$$ in explicit algebraic form.

Numerator:

$$$ A+B \;=\; x \;+\; \dfrac{2x}{1-x^{2}} = x\!\left(1\right) + x\!\left(\dfrac{2}{1-x^{2}}\right) = x\!\left(\dfrac{1-x^{2}}{1-x^{2}}\right) + x\!\left(\dfrac{2}{1-x^{2}}\right) = \dfrac{x\!\left(1-x^{2} + 2\right)}{1-x^{2}} = \dfrac{x\!\left(3 - x^{2}\right)}{1-x^{2}}. $$$

Denominator:

$$$ 1-AB = 1 - x\!\left(\dfrac{2x}{1-x^{2}}\right) = 1 - \dfrac{2x^{2}}{1-x^{2}} = \dfrac{1-x^{2}}{1-x^{2}} - \dfrac{2x^{2}}{1-x^{2}} = \dfrac{1 - x^{2} - 2x^{2}}{1-x^{2}} = \dfrac{1 - 3x^{2}}{1-x^{2}}. $$$

Putting numerator and denominator together:

$$$ \dfrac{A+B}{1-AB} = \dfrac{\dfrac{x(3 - x^{2})}{1-x^{2}}}{\dfrac{1 - 3x^{2}}{1-x^{2}}} = \dfrac{x(3 - x^{2})}{1 - 3x^{2}}, $$$

because the common factor $$1-x^{2}$$ cancels out completely.

Hence we have obtained

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

Since the inverse-tangent function is one-to-one in its principal branch, the arguments must be equal, and therefore

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

This matches Option B in the list provided.

Hence, the correct answer is Option B.