Example 1. (47th Putnam Competition) Find and prove the maximum value of $f(x)=x^{3}-3 x$, where $x$ is a real number satisfying $x^{4}+36 \leqslant 13 x^{2}$.

保留源文本的换行和格式，翻译结果如下：

Example 1. (47th Putnam Competition) Find and prove the maximum value of $f(x)=x^{3}-3 x$, where $x$ is a real number satisfying $x^{4}+36 \leqslant 13 x^{2}$.
From $x^{4}+36 \leqslant 13 x^{2}$, we can get
$$
-3 \leqslant x \leqslant-2,2 \leqslant x \leqslant 3 \text {. }
$$

Let $x_{i}3$. Then,
$$
\begin{array}{l}
f\left(x_{1}\right)-f\left(x_{2}\right) \\
=\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}-3\right)<0, \\
f\left(x_{1}\right)<f\left(x_{2}\right) .
\end{array}
$$

This shows that $f(x)$ is an increasing function on $[-3,-2]$ and $[2,3]$, and the maximum value is at the right endpoint of the interval. Therefore,
$$
\max f(x)=\max \{f(-2), f(3)\}=18
$$