$\frac{d}{dx}\left\{{\cos }^{-1}(\frac{4x^3}{64}-\frac{3}{4}x)\right\}=$

$x=4\cos \theta \Rightarrow \cos \theta =\frac{x}{4}\Rightarrow \theta ={\cos }^{-1}\left(\frac{x}{4}\right)$

$\frac{d}{dx}\left\{{\cos }^{-1}(\frac{4x^3}{64}-\frac{3}{4}x)\right\}$

$=\frac{d}{dx}\left\{{\cos }^{-1}\left(\cos 3\theta \right)\right\}$

$=\frac{d}{dx}\left(3\theta \right)$

$=3.\frac{d}{dx}\left(\theta \right)$

$=3.\frac{d}{dx}\left({\cos }^{-1}\left(\frac{x}{4}\right)\right)$

$=3.\frac{-1}{\sqrt{1-\frac{x^2}{16}}}\times \frac{1}{4}$

$=\frac{-3}{\sqrt{16-x^2}}$