The greatest value of the function $f(x)=\sin ^{-1} x^2$ in the interval $[-1 / \sqrt{2}, 1 / \sqrt{2}]$, is

$\frac{\pi}{6}$

We have,

$f(x)=\sin ^{-1} x^2 \Rightarrow f^{\prime}(x)=\frac{2 x}{\sqrt{1-x^4}}$

∴  $f^{\prime}(x)=0 \Rightarrow x=0$

Now,

$f\left(-\frac{1}{\sqrt{2}}\right)=\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}, f\left(\frac{1}{\sqrt{2}}\right)=\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}$

and, $f(0)=\sin ^{-1}(0)=0$.

Hence, the greatest value is $\frac{\pi}{6}$.