Step 1: Approach
Substitute $x=\tan\theta$ so that each expression turns into a double-angle formula.
Step 2: Substitution
$\dfrac{2x}{1+x^2}=\sin2\theta$, $\dfrac{1-x^2}{1+x^2}=\cos2\theta$, $\dfrac{2x}{1-x^2}=\tan2\theta$. For small positive $\theta$ (so $2\theta$ stays in the principal ranges) each inverse function returns $2\theta$.
Step 3: Equation
$6\theta-8\theta+4\theta=\dfrac\pi3$, so $2\theta=\dfrac\pi3$ and $\theta=\dfrac\pi6$.
Step 4: Answer
$x=\tan\dfrac\pi6=\dfrac1{\sqrt3}$, option (C).
Final Answer:
The equation reduces to $2\tan^{-1}x=\pi/3$, so $x=1/\sqrt3$, option (C).
\[ \boxed{\frac{1}{\sqrt{3}}} \]