Example 27 (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 $f(x, y, z) = x(1-y^2)(1-z^2) + y(1-z^2)(1-x^2) + z(1-x^2)(1-y^2)$.
From the given condition, by analogy with the structure of the trigonometric formula $\cot \alpha \cdot \cot \beta + \cot \beta \cdot \cot \gamma + \cot \gamma \cdot \cot \alpha = 1$. (*) Thus, we can let $x = \cot \alpha, y = \cot \beta, z = \cot \gamma$, where $\alpha, \beta, \gamma$ are acute angles and none of them are $\frac{\pi}{4}$. From the known conditions, we can derive (*) as follows:
$\tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \cdot \tan \beta \cdot \tan \gamma$.
That is, $\tan (\alpha + \beta) = -\tan \gamma = \tan (\pi - \gamma)$,
Thus, we have $\alpha + \beta = \pi - \gamma$, which means $\alpha + \beta + \gamma = \pi$.
Therefore, $\tan 2\alpha + \tan 2\beta + \tan 2\gamma = \tan 2\alpha \cdot \tan 2\beta \cdot \tan 2\gamma$.

So, $f(x, y, z) = \left(1 - x^2\right)\left(1 - y^2\right)\left(1 - z^2\right)\left(\frac{x}{1 - x^2} + \frac{y}{1 - y^2} + \frac{z}{1 - z^2}\right)$
$$
\begin{array}{l}
= \left(1 - \cot^2 \alpha\right)\left(1 - \cot^2 \beta\right)\left(1 - \cot^2 \gamma\right)\left(\frac{\cot \alpha}{1 - \cot^2 \alpha} + \frac{\cot \beta}{1 - \cot^2 \beta} + \frac{\cot \gamma}{1 - \cot^2 \gamma}\right) \\
= \frac{\tan^2 \alpha - 1}{\tan^2 \alpha} \cdot \frac{\tan^2 \beta - 1}{\tan^2 \beta} \cdot \frac{\tan^2 \gamma - 1}{\tan^2 \gamma}\left(\frac{-\tan \alpha}{1 - \tan^2 \alpha} + \frac{-\tan \beta}{1 - \tan^2 \beta} + \frac{-\tan \gamma}{1 - \tan^2 \gamma}\right) \\
= \frac{4}{\tan \alpha \cdot \tan \beta \cdot \tan \gamma} \cdot \frac{1 - \tan^2 \alpha}{2 \tan \alpha} \cdot \frac{1 - \tan^2 \beta}{2 \tan \beta} \cdot \frac{1 - \tan^2 \gamma}{2 \tan \gamma}\left(\frac{2 \tan \alpha}{1 - \tan^2 \alpha} + \frac{2 \tan \beta}{1 - \tan^2 \beta} + \frac{2 \tan \gamma}{1 - \tan^2 \gamma}\right) \\
= \frac{4}{\tan \alpha \cdot \tan \beta \cdot \tan \gamma} \cdot \cot 2\alpha \cdot \cot 2\beta + \cot 2\gamma \cdot (\tan 2\alpha + \tan 2\beta + \tan 2\gamma) \\
= \frac{4}{\tan \alpha \cdot \tan \beta \cdot \tan \gamma} \cdot \cot 2\alpha \cdot \cot 2\beta \cdot \cot 2\gamma \cdot \tan 2\alpha \cdot \tan 2\beta \cdot \tan 2\gamma \\
= \frac{4}{\tan \alpha \cdot \tan \beta \cdot \tan \gamma}
\end{array}
$$

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, $f(x, y, z) \leqslant \frac{4}{3 \sqrt{3}} = \frac{4 \sqrt{3}}{9}$.
Therefore, when and only when $\tan \alpha = \tan \beta = \tan \gamma$, i.e., $x = y = z = \frac{\sqrt{3}}{3}$, $f(x, y, z)$ achieves its maximum value $\frac{4}{9} \sqrt{3}$.