jmc

Let $a$ and $b$ be real numbers. Then $(a-b{)}^2\ge 0,$ which is equivalent to $$ab\le \frac{a^2+b^2}{2}.$$(This looks like AM-GM, but here, we want to show that it holds for all real numbers, not just nonnegative real numbers.)

Setting $a=x$ and $b=\sqrt{1-y^2},$ we get $$x\sqrt{1-y^2}\le \frac{x^2+1-y^2}{2}.$$Setting $a=y$ and $b=\sqrt{1-x^2},$ we get $$y\sqrt{1-x^2}\le \frac{y^2+1-x^2}{2}.$$Therefore, $$x\sqrt{1-y^2}+y\sqrt{1-x^2}\le \frac{x^2+1-y^2}{2}+\frac{y^2+1-x^2}{2}=1.$$Since $f(1,0)=1,$ the maximum value is $\boxed{1}.$