II. (16 points) Let $x, y \in \mathbf{R}$, and $x^{2}+y^{2}=2$, $|x| \neq|y|$. Find the minimum value of $\frac{1}{(x+y)^{2}}+\frac{1}{(x-y)^{2}}$.
$$
\begin{array}{l}
\text { II. Since } x^{2}+y^{2}=2 \text {, then, } \\
(x+y)^{2}+(x-y)^{2}=4 . \\
\text { Therefore, } \frac{1}{(x+y)^{2}}+\frac{1}{(x-y)^{2}} \\
=\frac{1}{4}\left(\frac{1}{(x+y)^{2}}+\frac{1}{(x-y)^{2}}\right)\left((x+y)^{2}+(x-y)^{2}\right) \\
\geqslant \frac{1}{4}(1+1)^{2}=1 .
\end{array}
$$

When $x=\sqrt{2}, y=0$, the above expression achieves its minimum value of 1.