Given function is:

$f(x) = \log |\cos x|$

Differentiating $f(x)$:

$f'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$

For interval $\left(0, \dfrac{\pi}{2}\right)$:

When $x \in \left(0, \dfrac{\pi}{2}\right)$, we have $x$ in the first quadrant.

Therefore, $\sin x > 0$ and $\cos x > 0$, which gives:

$\tan x = \frac{\sin x}{\cos x} > 0$

Thus, $f'(x) = -\tan x < 0$ for all $x \in \left(0, \dfrac{\pi}{2}\right)$.

Since $f'(x) < 0$, the function $f$ is decreasing on $\left(0, \dfrac{\pi}{2}\right)$.

For interval $\left(\dfrac{3\pi}{2}, 2\pi\right)$:

When $x \in \left(\dfrac{3\pi}{2}, 2\pi\right)$, we have $x$ in the fourth quadrant.

Therefore, $\sin x < 0$ and $\cos x > 0$, which gives:

$\tan x = \frac{\sin x}{\cos x} < 0$

Thus, $f'(x) = -\tan x > 0$ for all $x \in \left(\dfrac{3\pi}{2}, 2\pi\right)$.

Since $f'(x) > 0$, the function $f$ is increasing on $\left(\dfrac{3\pi}{2}, 2\pi\right)$.

∴ Hence proved.