$1 \cdot 31$ If all nonzero numbers $x$ and $y$ satisfy $x=\frac{1}{y}$, then $\left(x-\frac{1}{x}\right)\left(y+\frac{1}{y}\right)$ equals
(A) $2 x^{2}$.
(B) $2 y^{2}$.
(C) $x^{2}+y^{2}$.
(D) $x^{2}-y^{2}$.
(E) $y^{2}-x^{2}$.
(29th American High School Mathematics Examination, 1978)
[Solution] Let $x=\frac{1}{y}$, then $y=\frac{1}{x}$. Therefore,
$$
\left(x-\frac{1}{x}\right)\left(y+\frac{1}{y}\right)=(x-y)(x+y)=x^{2}-y^{2} \text{. }
$$

Hence, the correct choice is $(D)$.