The second order derivative of $\cos^{-1}(4x^3 - 3x)$ with respect to $\cos^{-1}(2x^2 - 1)$, where $\dfrac{1}{2} < x < 1$ is

Let $u = \cos^{-1}(4x^3 - 3x)$ and $v = \cos^{-1}(2x^2 - 1)$.

Substitute $x = \cos \theta$. Since $\dfrac{1}{2} < x < 1$, we have $0 < \theta < \dfrac{\pi}{3}$.

For $u$:
$u = \cos^{-1}(4\cos^3 \theta - 3\cos \theta) = \cos^{-1}(\cos 3\theta)$
Since $0 < \theta < \dfrac{\pi}{3}$, we get $0 < 3\theta < \pi$. In this interval, $\cos^{-1}(\cos 3\theta) = 3\theta$.
Thus, $u = 3\theta$.

For $v$:
$v = \cos^{-1}(2\cos^2 \theta - 1) = \cos^{-1}(\cos 2\theta)$
Since $0 < \theta < \dfrac{\pi}{3}$, we get $0 < 2\theta < \dfrac{2\pi}{3}$. In this interval, $\cos^{-1}(\cos 2\theta) = 2\theta$.
Thus, $v = 2\theta$.

We can express $u$ in terms of $v$ as:
$u = \dfrac{3}{2} v$

Differentiating $u$ with respect to $v$, we get the first order derivative:
$\dfrac{du}{dv} = \dfrac{3}{2}$

Differentiating again with respect to $v$, we get the second order derivative:
$\dfrac{d^2u}{dv^2} = \dfrac{d}{dv}\left(\dfrac{3}{2}\right) = 0$

Answer: $0$