Example 2 If $x$, $y$, $z$ are all real numbers, and $x^{2}+y^{2}+z^{2}$ $=1$, then the maximum value of $\sqrt{2} x y+y z$ is $\qquad$
Solve: Given $x^{2}+y^{2}+z^{2}=1$, let
$$
\begin{array}{l}
x=\cos \theta \cdot \cos \varphi, \\
z=\cos \theta \cdot \sin \varphi, \\
y=\sin \theta \quad (\theta, \varphi \in [0, 2\pi)) . \\
\text{Then } \sqrt{2} x y + y z \\
=\sin \theta \cdot \cos \theta \cdot (\sqrt{2} \cos \varphi + \sin \varphi) \\
=\frac{1}{2} \sin 2 \theta \cdot (\sqrt{2} \cos \varphi + \sin \varphi) \\
\leqslant \frac{1}{2} \times 1 \times \sqrt{3},
\end{array}
$$

Equality holds if and only if $\sin 2 \theta = 1, \sqrt{2} \cos \varphi + \sin \varphi = \sqrt{3}$, i.e.,
$$
x=\frac{\sqrt{3}}{3}, y=\frac{\sqrt{2}}{2}, z=\frac{\sqrt{6}}{6}
$$

Thus, the maximum value of $\sqrt{2} x y + y z$ is $\frac{\sqrt{3}}{2}$.