9. Let $\min \{a, b\}$ be the minimum of the two numbers $a$ and $b$. When positive numbers $x$ and $y$ vary, $t=\min \left\{x, \frac{y}{x^{2}+y^{2}}\right\}$ also varies. Then the maximum value of $t$ is
9. $\frac{\sqrt{2}}{2}$.
$t^{2} \leqslant x \cdot \frac{y}{x^{2}+y^{2}} \leqslant \frac{x y}{2 x y}=\frac{1}{2}$, so $t \leqslant \frac{\sqrt{2}}{2}$, with equality if and only if $x=y=\frac{\sqrt{2}}{2}$.