For a continuous function 

$f:\left[0,1\right]⟶\left[0,1\right]$ it is given that $x^2+{\left(f\left(x\right)\right)}^2\le 1$

$\Rightarrow {\left(f\left(x\right)\right)}^2\le 1-x^2$

$\Rightarrow f\left(x\right)\le \sqrt{1-x^2} \left\{\because f\left(x\right)\in \left[0,1\right]\right\}$

$\therefore {\int }_0^{1}f\left(x\right)dx\le {\int }_0^1\sqrt{1-x^2}dx$

$\Rightarrow {\int }_0^{1}f\left(x\right)dx\le {\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x\right]}_0^1$

$\Rightarrow {\int }_0^{1}f\left(x\right)dx\le \frac{\pi }{4}$

So, $f\left(x\right)=\sqrt{1-x^2}$, because it is given that ${\int }_0^{1}f\left(x\right)=\frac{\pi }{4}$.

Now, ${\int }_{1/2}^{1\sqrt{2}}\frac{f\left(x\right)}{1-x^2}dx={\int }_{1/2}^{1\sqrt{2}}\frac{dx}{\sqrt{1-x^2}}={\left[{\sin }^{-1}x\right]}_{1/2}^{1/\sqrt{2}}$

$=\frac{\pi }{4}-\frac{\pi }{6}=\frac{\pi }{12}$