Example 4 Let $x \geqslant 0, y \geqslant 0, x^{2}+\frac{y^{2}}{2}=1$, then the maximum value of $x \sqrt{1+y^{2}}$ is $\qquad$
Let $\left\{\begin{array}{l}x=\cos \theta, \\ y=\sqrt{2} \sin \theta\end{array}\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)\right.$.
Then
$$\begin{array}{c}
x \sqrt{1+y^{2}}=\cos \theta \sqrt{1+2 \sin ^{2} \theta}= \\
\sqrt{2 \cos ^{2} \theta\left(1+2 \sin ^{2} \theta\right) \cdot \frac{1}{2}} \leqslant \\
\sqrt{\frac{1}{2} \cdot\left[\frac{2 \cos ^{2} \theta+\left(1+2 \sin ^{2} \theta\right)}{2}\right]^{2}}=\frac{3 \sqrt{2}}{4}
\end{array}$$

When $2 \cos ^{2} \theta=1+2 \sin ^{2} \theta$, i.e., $\theta=\frac{\pi}{6}$, then $x=\frac{\sqrt{3}}{2}, y=\frac{\sqrt{2}}{2}$, and $x \sqrt{1+y^{2}}$ achieves its maximum value $\frac{3 \sqrt{2}}{4}$.