5. Given that $x$ and $y$ 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
A. $\frac{8}{5}$
B. $\frac{24}{11}$
C. $\frac{12}{7}$
D. $\frac{12}{5}$
$$
u=\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}}=\frac{4}{4-x^{2}}+\frac{9 x^{2}}{9 x^{2}-1}=\frac{36}{36-9 x^{2}}+\frac{1}{9 x^{2}-1}+1
$$
\geqslant \frac{(6+1)^{2}}{35}+1=\frac{12}{5}, equality holds when \frac{6}{36-9 x^{2}}=\frac{1}{9 x^{2}-1} \Rightarrow x^{2}=\frac{2}{3}, y^{2}=\frac{3}{2}.
Thus, the minimum value of the function \( u=\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}} \) is \(\frac{12}{5}\). Therefore, the answer is \(D\).