8. Given $x, y \in(-\sqrt{2}, \sqrt{2})$, and $x y=1$. Then the minimum value of $\frac{2}{2-x^{2}}+\frac{4}{4-y^{2}}$ is ( ).
(A) $\frac{20}{7}$
(B) $\frac{12}{7}$
(C) $\frac{16+4 \sqrt{2}}{7}$
(D) $\frac{16-4 \sqrt{2}}{7}$
8. C.

From the given, we have $y=\frac{1}{x}$. Therefore,
$$
\begin{array}{l}
\frac{2}{2-x^{2}}+\frac{4}{4-y^{2}}=\frac{2}{2-x^{2}}+\frac{4}{4-\frac{1}{x^{2}}} \\
=\frac{-4 x^{4}+16 x^{2}-2}{-4 x^{4}+9 x^{2}-2}=1+\frac{7 x^{2}}{-4 x^{4}+9 x^{2}-2} \\
=1+\frac{7}{9-\left(4 x^{2}+\frac{2}{x^{2}}\right)} .
\end{array}
$$

Since $4 x^{2}+\frac{2}{x^{2}} \geqslant 4 \sqrt{2}$, with equality if and only if $4 x^{2}=\frac{2}{x^{2}}$, i.e., $x=$ $\frac{\sqrt{8}}{2}$. Therefore, when $x=\frac{\sqrt[4]{8}}{2}$, $\frac{2}{2-x^{2}}+\frac{4}{4-y^{2}}$ achieves its minimum value $\frac{16+4 \sqrt{2}}{7}$.