Example 5 If real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$, find the maximum value of $\left(x^{2}-y z\right)\left(y^{2}-z x\right)\left(z^{2}-x y\right)$.
(1) $0 \in\{x, y, z\}$, without loss of generality, let $z=0$.
$$
\text { Original expression }=-x^{3} y^{3} \leqslant \frac{\left(x^{2}+y^{2}\right)^{3}}{8}=\frac{1}{8} \text {, }
$$

Equality holds if and only if $x=\frac{\sqrt{2}}{2}, y=-\frac{\sqrt{2}}{2}, z=0$ and their cyclic permutations.
(2) $0 \notin\{x, y, z\}$, without loss of generality, let $xy z$, it is easy to see that the maximum value is achieved at this point.
Differentiate the original expression with respect to variable $z$:
$$
\begin{array}{l}
f^{\prime}(z)=4 x y z^{3}-3\left(x^{3}+y^{3}\right) z^{2}+x^{4} y+x y^{4}, \\
f^{\prime \prime}(z)=12 x y z^{2}-6\left(x^{3}+y^{3}\right) z .
\end{array}
$$

Then, when $z=\frac{x^{3}+y^{3}}{2 x y}$, $f^{\prime}(z)$ reaches its minimum value.
At this point, $4 x^{3} y^{3}0 \text {. }
$$

Therefore, the original expression reaches its maximum value when $z=y$. At this point,
$$
\begin{array}{l}
\text { Original expression }=\left(x^{2}-y^{2}\right)\left(y^{2}-x y\right)^{2} \\
=\left(1-3 y^{2}\right) y^{2}\left(1-y^{2}-2 x y\right) . \\
\text { Let } y^{2}=m<\frac{1}{3} .
\end{array}
$$

Then the original expression $=(1-3 m) m\left(1-m-2 \sqrt{m-2 m^{2}}\right)$ $<m(1-2 m)^{2}=g(m)$.
Thus, $g^{\prime}(m)=12 m^{2}-8 m+1$.
It is easy to see that $\frac{1}{2} 、 \frac{1}{6}$ are the two roots of $g^{\prime}(m)=0$.
Therefore, $g(m) \leqslant g\left(\frac{1}{6}\right)=\frac{2}{27}<\frac{1}{8}$.
In summary, the maximum value of the original expression is $\frac{1}{8}$.