Example 5 Let $x, y, z \in \mathbf{R}^{+}$, and $x^{4}+y^{4}+z^{4}=1$, find
$$f(x, y, z)=\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}}$$

the minimum value.
Solve by transforming the original expression into
$$f(x, y, z)=\frac{x^{4}}{x\left(1-x^{8}\right)}+\frac{y^{4}}{y\left(1-y^{8}\right)}+\frac{z^{4}}{z\left(1-z^{8}\right)}$$

For $w \in(0,1)$, let $\phi(w)=w\left(1-w^{8}\right)$, and first find the maximum value of $\phi(w)$. Choose a parameter $a$, and use $G_{9} \leqslant A_{9}$, to get
$$\begin{aligned}
a(\phi(w))^{8} & =a w^{8}\left(1-w^{8}\right)^{8} \\
& \leqslant\left[\frac{1}{9}\left(a w^{8}+8\left(1-w^{8}\right)\right)\right]^{9} \\
& =\left[\frac{1}{9}\left(8+(a-8) w^{8}\right)\right]^{9}
\end{aligned}$$

Taking $a=8$, we get
$$8(\phi(w))^{8} \leqslant\left(\frac{8}{9}\right)^{9}$$

Since $\phi(w)>0$, it follows that
$$\phi(w) \leqslant \frac{8}{\sqrt[4]{3^{9}}}$$

Thus,
$$f(x, y, z) \geqslant \frac{x^{4}+y^{4}+z^{4}}{8} \cdot \sqrt[4]{3^{9}}=\frac{9}{8} \sqrt[4]{3},$$

with equality holding when $x=y=z=\frac{1}{\sqrt[4]{3}}$.
Therefore, the minimum value of $f(x, y, z)$ is $\frac{9}{8} \sqrt[4]{3}$.