Differentiate $\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)$ with respect to $\tan^{-1}\left(\dfrac{3x-x^3}{1-3x^2}\right)$
$\left[\begin{aligned}
\text{Note: } & \tan 3\theta = \dfrac{3 \tan \theta - \tan^3 \theta}{1-3\tan^2 \theta} \\
& \cos 2\theta = \dfrac{1-\tan^2 \theta}{1+\tan^2 \theta}
\end{aligned}\right]$
Let $u = \cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)$, $v = \tan^{-1}\left(\dfrac{3x-x^3}{1-3x^2}\right)$
Then $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} \;\; \cdots$ (1)
Let $x=\tan \alpha \;\; \cdots$ (2)
Differentiating equation(2) with respect to $\alpha$ gives
$\dfrac{dx}{d\alpha} = \sec^2 \alpha \;\; \cdots$ (3)
In view of equation (2), u becomes
$\begin{aligned}
u & = \cos^{-1}\left(\dfrac{1-\tan^2 \alpha}{1+\tan^2 \alpha}\right) \\
& = \cos^{-1} \left(\cos 2 \alpha\right) \\
& = 2\alpha
\end{aligned}$
$\therefore$ $\;$ $\dfrac{du}{d\alpha} = 2$ $\; \cdots$ (4)
In view of equation (2), v becomes
$\begin{aligned}
v & = \tan^{-1}\left(\dfrac{3\tan \alpha - \tan^3 \alpha}{1-3\tan^2 \alpha}\right) \\
& = \tan^{-1} \left(\tan 3 \alpha\right) \\
& = 3\alpha
\end{aligned}$
$\therefore$ $\;$ $\dfrac{dv}{d\alpha} = 3$ $\; \cdots$ (5)
Now, $\dfrac{du}{dx} = \dfrac{du / d\alpha}{dx / d\alpha} = \dfrac{2}{\sec^2 \alpha}$ $\;\; \cdots$ (6) [from equations (3) and (4)]
and $\dfrac{dv}{dx} = \dfrac{dv / d\alpha}{dx / d\alpha} = \dfrac{3}{\sec^2 \alpha}$ $\;\; \cdots$ (7) [from equations (3) and (5)]
$\therefore$ $\;$ In view of equations (6) and (7) equation (1) becomes
$\dfrac{du}{dv}= \dfrac{2 / \sec^2 \alpha}{3 / \sec^2 \alpha} = \dfrac{2}{3}$