Solve this

$y=\cos ^{-1}\left\{\frac{2 x-3 \sqrt{1-x^{2}}}{\sqrt{13}}\right\}$

Put $x=\cos \theta$

$y=\cos ^{-1}\left\{\frac{2 \cos \theta-3 \sqrt{1-\cos ^{2} \theta}}{\sqrt{13}}\right\}$

$y=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos \theta-\frac{3}{\sqrt{13}} \sin \theta\right\}$

let $\cos \phi=\frac{2}{\sqrt{13}}$

Now,

$\Rightarrow \sin ^{2} \phi=1-\cos ^{2} \phi$

$\Rightarrow \sin \phi=\sqrt{1-\cos ^{2} \phi}$

$\Rightarrow \sin \phi=\sqrt{1-\frac{4}{13}}$

$\Rightarrow \sin \phi=\frac{3}{\sqrt{13}}$

Again,

$y=\cos ^{-1}\{\cos \phi \cos \theta-\sin \phi \sin \theta\}$

Using $\cos A \cos B-\sin A \sin B=\cos (A+B)$

$y=\cos ^{-1}\{\cos (\phi+\theta)\}$

$y=\phi+\theta$

$y=\cos ^{-1}\left\{\frac{2}{\sqrt{13}}\right\}+\cos ^{-1} x$

Differentiating w.r.t $\mathrm{x}$, we get

$\frac{d y}{d x}=\frac{d}{d x}\left(\cos ^{-1}\left\{\frac{2}{\sqrt{13}}\right\}+\cos ^{-1} x\right)$

$\frac{d y}{d x}=0+\left(-\frac{1}{\sqrt{1-x^{2}}}\right)$

$\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}}$