Example 4 It is known that for any $x, y, z \geqslant 0$, we have
$$
\begin{array}{l}
x^{3}+y^{3}+z^{3}-3 x y z \\
\geqslant c|(x-y)(y-z)(z-x)| .
\end{array}
$$

Find the maximum value of $c$.
【Analysis】Left side of equation (1)
$$
\begin{array}{l}
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \\
=(x+y+z) \frac{(x-z)^{2}+(y-z)^{2}+(z-x)^{2}}{2} .
\end{array}
$$

Assume without loss of generality that $x \geqslant y \geqslant z \geqslant 0$.
If $x=y$ or $y=z$, inequality (1) holds for any real number $c$.

If $x>y>z \geqslant 0$, fixing $x-y$ and $y-z$, then $z-x$ is also fixed. To maximize $c$, the left side of inequality (1) should be minimized, i.e., $x+y+z$ should be minimized.

Since $x+y+z=(x-y)+2(y-z)+3 z$, the left side of inequality (1) is minimized when $z=0$.
Now consider the case $x>y>z=0$.
Inequality (1) becomes $x^{3}+y^{3} \geqslant c x y(x-y)$.
Let $t=\frac{x}{y}(t>1)$. Then $t^{3}+1 \geqslant c t(t-1)$.
Thus, $c_{\max }=f_{\text {min }}(t)\left(f(t)=\frac{t^{3}+1}{t(t-1)}\right)$.
Therefore, $f^{\prime}(t)=\frac{t^{4}-2 t^{3}-2 t+1}{t^{2}(t-1)^{2}}$.
$$
\begin{array}{l}
\text { and } f^{\prime}(t)=0 \\
\Leftrightarrow t^{4}-2 t^{3}-2 t+1=0 \\
\Leftrightarrow\left(t+\frac{1}{t}\right)^{2}-2\left(t+\frac{1}{t}\right)-2=0 \\
\Leftrightarrow t+\frac{1}{t}=1+\sqrt{3}\left(t+\frac{1}{t}=1-\sqrt{3} \text { is discarded }\right) \\
\Leftrightarrow t=\frac{1+\sqrt{3}+\sqrt{2 \sqrt{3}}}{2}\left(t=\frac{1+\sqrt{3}-\sqrt{2 \sqrt{3}}}{2}\right.
\end{array}
$$

is discarded).
When $t \in\left(1, \frac{1+\sqrt{3}+\sqrt{2 \sqrt{3}}}{2}\right)$, $f^{\prime}(t)0$, $f(t)$ is strictly monotonically increasing.
$$
\begin{array}{l}
\text { then } c_{\text {max }}=f_{\text {min }}(t)=f\left(\frac{1+\sqrt{3}+\sqrt{2 \sqrt{3}}}{2}\right) \\
=\left(\frac{\sqrt{6}+3 \sqrt{2}}{2}\right) \sqrt[4]{3} .
\end{array}
$$