The range of $x$ for which the equation $\sin ^{-1}\left(\dfrac{2 x}{1+x^2}\right)=2 \tan ^{-1} x$ holds true

The expression $\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$ is defined for $\left|\dfrac{2x}{1+x^2}\right| \leq 1$, which implies $2|x| \leq 1+x^2$, or $(|x|-1)^2 \geq 0$, which is true for all $x \in R$.

Recall the identity $\sin^{-1}\left(\dfrac{2x}{1+x^2}\right) = 2 \tan^{-1} x$ for $|x| \leq 1$.

For $x > 1$, let $x = \tan \theta$ where $\theta \in (\pi/4, \pi/2)$. Then $\dfrac{2x}{1+x^2} = \sin(2\theta)$. Since $2\theta \in (\pi/2, \pi)$, $\sin^{-1}(\sin(2\theta)) = \pi - 2\theta = \pi - 2\tan^{-1} x$. Thus, the identity does not hold for $x > 1$.

For $x < -1$, let $x = \tan \theta$ where $\theta \in (-\pi/2, -\pi/4)$. Then $2\theta \in (-\pi, -\pi/2)$. $\sin^{-1}(\sin(2\theta)) = -\pi - 2\theta = -\pi - 2\tan^{-1} x$. Thus, the identity does not hold for $x < -1$.

Therefore, the equation holds true if and only if $|x| \leq 1$.

Answer: $|x| \leq 1$