Example 8 Given that $x, y, z$ are non-negative real numbers. If for any $x, y, z$, we have
$$
\begin{array}{l}
x^{3}+y^{3}+z^{3}-3 x y z \\
\geqslant k|(x-y)(y-z)(z-x)|,
\end{array}
$$

find the maximum value of $k$.
Organizing we get
$$
2 k^{3}-9 k^{2}-27=2 k(k-3) \sqrt{k(k-3)} \text {. }
$$

Squaring both sides and organizing we get
$$
\begin{array}{l}
k^{4}-18 k^{2}-27=0 \\
\Rightarrow k^{2}=9+6 \sqrt{3} \\
\Rightarrow k=\sqrt{9+6 \sqrt{3}} .
\end{array}
$$

When $x=0, \frac{z}{y}=\frac{k-3+\sqrt{k^{2}-3 k}}{3}$, the equality in the original inequality holds.
Solution: By symmetry, assume $x \leqslant y \leqslant z$. Then
$$
\begin{array}{l}
(x+y+z)\left((x-y)^{2}+(y-z)^{2}+\right. \\
\left.(z-x)^{2}\right)-k(x-y)(y-z)(z-x) \geqslant 0 .
\end{array}
$$

Let $y=x+a, z=x+a+b, a, b \geqslant 0$. Then
$$
\begin{array}{l}
(2 a+b+3 x)\left(a^{2}+b^{2}+(a+b)^{2}\right) \\
\geqslant 2 a b(a+b) k .
\end{array}
$$

Since $x \geqslant 0$, it suffices to have
$$
\begin{array}{l}
(2 a+b)\left(a^{2}+b^{2}+(a+b)^{2}\right) \\
\geqslant 2 a b(a+b) k,
\end{array}
$$

and if $a=b=0$, the inequality clearly holds, i.e.,
$$
\begin{array}{l}
b^{3}+(3-k) b^{2} a+(3-k) b a^{2}+2 a^{3} \geqslant 0 \\
\Leftrightarrow\left(\frac{b}{a}\right)^{3}+(3-k)\left(\frac{b}{a}\right)^{2}+(3-k)\left(\frac{b}{a}\right)+2 \geqslant 0 . \\
\text { Let } t=\frac{b}{a}>0, \\
f(t)=t^{3}+(3-k) t^{2}+(3-k) t+2 . \\
\text { Then } f^{\prime}(t)=3 t^{2}+(6-2 k) t+(3-k), \\
f^{\prime}(t)=0 \Leftrightarrow t=\frac{k-3+\sqrt{k^{2}-3 k}}{3} . \\
\text { Hence } f\left(\frac{k-3+\sqrt{k^{2}-3 k}}{3}\right)=0 .
\end{array}
$$