3. Given $x, y$ are in the interval $(-2,2)$, and $x y=-1$, then the minimum value of the function $u=\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}}$ is
3. $\frac{12}{7}$
$$
\text { Detailed explanation: } \begin{aligned}
u & =\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}}=\frac{36-4 y^{2}+36-9 x^{2}}{36-9 x^{2}-4 y^{2}+x^{2} y^{2}} \\
& =1+\frac{35}{37-\left(4 x^{2}+9 y^{2}\right)} \geqslant 1+\frac{35}{37+12}=\frac{12}{7} .
\end{aligned}
$$