9. (15 points) Given a non-zero real number $a$, solve the system of equations for $x$ and $y$
$$
\left\{\begin{array}{l}
x y-\frac{x}{y}=a, \\
x y-\frac{y}{x}=\frac{1}{a} .
\end{array}\right.
$$
Sure, here is the translated text:

```
9. The original system of equations is
$$
\left\{\begin{array}{l}
x y-a=\frac{x}{y}, \\
x y-\frac{1}{a}=\frac{y}{x} .
\end{array}\right.
$$

Multiplying the two equations gives
$$
\begin{array}{l}
(x y-a)\left(x y-\frac{1}{a}\right)=1 \\
\Rightarrow(x y)^{2}-\left(a+\frac{1}{a}\right)(x y)=0 .
\end{array}
$$

Given that the system of equations is meaningful, we know $x y \neq 0$.
Thus, $x y=a+\frac{1}{a}$.
Substituting equation (2) into equation (1) gives
$$
\frac{x}{y}=\frac{1}{a} \Rightarrow y=a x \text {. }
$$

Substituting $y=a x$ into equation (2) gives
$$
a x^{2}=a+\frac{1}{a} \Rightarrow x= \pm \frac{\sqrt{a^{2}+1}}{a} \text {. }
$$

Therefore, $y=a x= \pm \sqrt{a^{2}+1}$.
Hence, the solutions to the system of equations are
$$
\begin{aligned}
(x, y)= & \left(\frac{\sqrt{a^{2}+1}}{a}, \sqrt{a^{2}+1}\right) \\
& \text { or }\left(-\frac{\sqrt{a^{2}+1}}{a},-\sqrt{a^{2}+1}\right) .
\end{aligned}
$$
```