9. Given real numbers $x, y$ satisfy $x^{2}+\frac{y^{2}}{16}=1$. Then the maximum value of $x \sqrt{2+y^{2}}$ is $\qquad$
9. $\frac{9}{4}$.

From the problem, we know $16 x^{2}+y^{2}=16$.
Let $s=x \cdot \sqrt{2+y^{2}}$. Then
$$
\begin{array}{l}
s^{2}=x^{2}\left(2+y^{2}\right)=\frac{1}{16} \times 16 x^{2}\left(2+y^{2}\right) \\
\leqslant \frac{1}{16}\left(\frac{16 x^{2}+2+y^{2}}{2}\right)^{2}=\frac{81}{16} .
\end{array}
$$

Therefore, $s \leqslant \frac{9}{4}$, and the equality holds when $x=\frac{3}{4}, y= \pm \sqrt{7}$.