Solution I We have
$$\begin{aligned} u &= \frac{4}{4-x^2} + \frac{9x^2}{9x^2-1} = 1 + \frac{35x^2}{-9x^4 + 37x^2 - 4} \\ &= 1 + \frac{35}{37 - \left( \left( 3x - \frac{2}{x} \right)^2 + 12 \right)} \end{aligned}$$
Since $x \in (-2, -\frac{1}{2}) \cup (\frac{1}{2}, 2)$, so $u$ reaches the minimum value $\frac{12}{5}$ when $x = \pm\sqrt{\frac{2}{3}}$. Answer: D.

Solution II It is known from the conditions that $4-x^2 > 0$ and $9-y^2 > 0$. Then
$$\begin{aligned} u &\ge 2\sqrt{\frac{4}{4-x^2} \cdot \frac{9}{9-y^2}} = \frac{12}{\sqrt{36-9x^2-4y^2+(xy)^2}} \\ &= \frac{12}{\sqrt{37-9x^2-4y^2}} \ge \frac{12}{\sqrt{37-2\sqrt{36(xy)^2}}} = \frac{12}{5}. \end{aligned}$$
Since $u$ is $\frac{12}{5}$ when $x = \sqrt{\frac{2}{3}}$ and $y = -\sqrt{\frac{3}{2}}$, so $u$ reaches the minimum. Answer: D.