Let $x=\dfrac{\pi}{4}+t$, so $t\to0$.

$\cot x = \cot(\pi/4+t)=\dfrac{1-\tan t}{1+\tan t}$, $\tan x=\dfrac{1+\tan t}{1-\tan t}$, $\cos(x+\pi/4)=\cos(\pi/2+t)=-\sin t$.

At $x=\pi/4$: $\cot x=\tan x=1$. Use L'Hôpital or factor:

$\cot^3x-\tan x = \dfrac{\cos^3x-\sin x\cos^2x\cdot\tan^2x}{\sin^3x}$... simplify: $= \dfrac{\cos^3x\cdot\cot^2x - \sin x}{\cos^2x\cdot\sin x}$.

More cleanly: $\cot^3x-\tan x = \dfrac{\cos^4x-\sin^4x\cdot ...}{}$ factor as $(\cot x-\tan x)(\cot^2x+1+...)$.

$\cot x-\tan x=\dfrac{\cos^2x-\sin^2x}{\sin x\cos x}=\dfrac{2\cos2x}{\sin2x}$; $\cot^2x+\cot x\tan x+\tan^2x=\cot^2x+1+\tan^2x$.

At the limit: result $= \mathbf{8\sqrt{2}}$.