Example 22 (Mathematical Problem 1434 from "Mathematical Bulletin") Let positive numbers $x, y, z$ be not equal to 1, and $xy + yz + zx = 1$. Try to find the maximum value of $f(x, y, z) = x(1-y^2)(1-z^2) + y(1-z^2)(1-x^2) + z(1-x^2)(1-y^2)$.
Let $x=\cot \alpha, y=\cot \beta, z=\cot \gamma$, where $\alpha, \beta, \gamma$ are acute angles and none of them is $\frac{\pi}{4}$. From the condition, we have $\cot \alpha \cdot \cot \beta + \cot \beta \cdot \cot \gamma + \cot \gamma \cdot \cot \alpha = 1$, which implies $\tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \cdot \tan \beta \cdot \tan \gamma$, or equivalently, $\tan (\alpha + \beta) = -\tan \gamma = \tan (\pi - \gamma)$,
thus $\alpha + \beta = \pi - \gamma$, which means $\alpha + \beta + \gamma = \pi$.
From this, we deduce that $\tan 2\alpha + \tan 2\beta + \tan 2\gamma = \tan 2\alpha \cdot \tan 2\beta \cdot \tan 2\gamma$.
Therefore,
\[
\begin{aligned}
f(x, y, z) & = \left(1 - x^2\right)\left(1 - y^2\right)\left(1 - z^2\right)\left(\frac{x}{1 - x^2} + \frac{y}{1 - y^2} + \frac{z}{1 - z^2}\right) \\
& = \left(1 - \cot^2 \alpha\right)\left(1 - \cot^2 \beta\right)\left(1 - \cot^2 \gamma\right) \cdot \left(\frac{\cot \alpha}{1 - \cot^2 \alpha} + \frac{\cot \beta}{1 - \cot^2 \beta} + \frac{\cot \gamma}{1 - \cot^2 \gamma}\right) \\
& = \frac{\tan^2 \alpha - 1}{\tan^2 \alpha} \cdot \frac{\tan^2 \beta - 1}{\tan^2 \beta} \cdot \frac{\tan^2 \gamma - 1}{\tan^2 \gamma} \cdot \left(\frac{-\tan \alpha}{1 - \tan^2 \alpha} + \frac{-\tan \beta}{1 - \tan^2 \beta} + \frac{-\tan \gamma}{1 - \tan^2 \gamma}\right) \\
& = \frac{4}{\tan \alpha \cdot \tan \beta \cdot \tan \gamma} \cdot \frac{1 - \tan^2 \alpha}{2 \tan \alpha} \cdot \frac{1 - \tan^2 \beta}{2 \tan \beta} \cdot \frac{1 - \tan^2 \gamma}{2 \tan \gamma} \cdot \left(\frac{2 \tan \alpha}{1 - \tan^2 \alpha} + \right. \\
& \left. = \frac{2 \tan \beta}{1 - \tan^2 \beta} + \frac{2 \tan \gamma}{1 - \tan^2 \gamma}\right) \\
& = \frac{4}{\tan \alpha \cdot \tan \beta \cdot \tan \gamma} \cdot \cot 2\alpha \cdot \cot 2\beta \cdot \cot 2\gamma \cdot (\tan 2\alpha + \tan 2\beta + \tan 2\gamma) \\
& = \frac{4}{\tan \alpha \cdot \tan \beta \cdot \tan \gamma}.
\end{aligned}
\]

Since $\tan \alpha \cdot \tan \beta \cdot \tan \gamma = \tan \alpha + \tan \beta + \tan \gamma \geqslant 3 \sqrt[3]{\tan \alpha \cdot \tan \beta \cdot \tan \gamma}$,
we have $\tan \alpha \cdot \tan \beta \cdot \tan \gamma \geqslant 3 \sqrt{3}$.
Thus, $f(x, y, z) \leqslant \frac{4}{3 \sqrt{3}} = \frac{4}{9} \sqrt{3}$.

The equality holds if and only if $\tan \alpha = \tan \beta = \tan \gamma$, i.e., $x = y = z = \frac{\sqrt{3}}{3}$.
Therefore, the maximum value of $f(x, y, z)$ is $\frac{4}{9} \sqrt{3}$.