if

$\cos ^{-1} x-\cos ^{-1} y=\frac{\pi}{6}$

$\Rightarrow \frac{\pi}{2}-\sin ^{-1} x-\frac{\pi}{2}+\sin ^{-1} y=\frac{\pi}{6} \quad\left[\because \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x\right]$

$\Rightarrow-\left(\sin ^{-1} x-\sin ^{-1} y\right)=\frac{\pi}{6}$

$\Rightarrow \sin ^{-1} x-\sin ^{-1} y=-\frac{\pi}{6}$

Solving $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{3}$ and $\sin ^{-1} x-\sin ^{-1} y=-\frac{\pi}{6}$, we will get

$2 \sin ^{-1} x=\frac{\pi}{6}$

$\Rightarrow \sin ^{-1} x=\frac{\pi}{12}$

$\Rightarrow x=\sin \frac{\pi}{12}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$

and

$\sin ^{-1} y=\frac{\pi}{3}-\sin ^{-1} x$

$\Rightarrow \sin ^{-1} y=\frac{\pi}{3}-\frac{\pi}{12}$

$\Rightarrow \sin ^{-1} y=\frac{\pi}{4}$

$\Rightarrow y=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}$