Write the value

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

Then, $\sin y=\frac{1}{3}$

Now, $\cos y=\sqrt{1-\sin ^{2} y}$

$\Rightarrow \cos y=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}$

$\cos \left(2 \sin ^{-1} \frac{1}{3}\right)=\cos (2 y)$

$=\cos ^{2} y-\sin ^{2} y \quad\left[\because \cos 2 x=\cos ^{2} x-\sin ^{2} x\right]$

$=\left(\frac{2 \sqrt{2}}{3}\right)^{2}-\left(\frac{1}{3}\right)^{2}$

$=\frac{8}{9}-\frac{1}{9}$

$=\frac{7}{9}$

$\therefore \cos \left(2 \sin ^{-1} \frac{1}{3}\right)=\frac{7}{9}$