9. Given that $x$ and $y$ are real numbers, satisfying $x^{2}+y^{2} \leqslant 1$. Then the maximum value of $\left|x^{2}+2 x y-y^{2}\right|$ is $\qquad$ .
9. $\sqrt{2}$.

Let $x=r \cos \theta, y=r \sin \theta(0 \leqslant r \leqslant 1,-\pi<\theta \leqslant \pi)$,
then
$$
\begin{aligned}
& \left|x^{2}+2 x y-y^{2}\right| \\
& =r^{2}\left|\cos ^{2} \theta+2 \sin \theta \cdot \cos \theta-\sin ^{2} \theta\right| \\
= & r^{2}|\sin 2 \theta+\cos 2 \theta| \\
= & \sqrt{2} r^{2}\left|\sin \left(2 \theta+\frac{\pi}{4}\right)\right| \leqslant \sqrt{2},
\end{aligned}
$$

The equality holds if and only if $r=1$, i.e., $2 \theta+\frac{\pi}{4}= \pm \frac{\pi}{2}$.