Example 2 Given that $x, y$ are both in the interval $(-2,2)$, and $x y=-1$, then what is the minimum value of the function $u=\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}}$?
Given $y=-\frac{1}{x}$, we have
$$
u=\frac{4}{4-x^{2}}+\frac{9 x^{2}}{9 x^{2}-1}=\frac{-9 x^{4}+72 x^{2}-4}{-9 x^{4}+37 x^{2}-4}=1+\frac{35}{37-\left(9 x^{2}+\frac{4}{x^{2}}\right)}
$$

Since $x \in(-2,0) \cup(0,2)$, when $9 x^{2}=\frac{4}{x^{2}}$, i.e., $x^{2}=\frac{2}{3}$, the value of $9 x^{2}+\frac{4}{x^{2}}$ is minimized, and at this point, the function $u$ has a minimum value of $\frac{12}{5}$.