Example 7 (2003 National High School Competition Question) Given that $x$ and $y$ are both in the interval $(-2,2)$, and $xy=-1$, then the minimum value of the function $u=\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}}$ is ( ).
A. $\frac{8}{5}$
B. $\frac{24}{11}$
C. $\frac{12}{7}$
D. $\frac{12}{5}$
Solution: Choose D. Reason: From the given, we have $y=-\frac{1}{x}$, thus
$$
u=\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}}=\frac{4}{4-x^{2}}+\frac{9 x^{2}}{9 x^{2}-1}=1+\frac{35}{37-\left(9 x^{2}+\frac{4}{x^{2}}\right)} .
$$

And $x \in\left(-2,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, 2\right)$, 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 the smallest, at this time $u$ has the minimum value $\frac{12}{5}$.