Consider the first equations of the system. It is easy to see that $xy \neq 0$, and we
$$\text{get } \frac{9}{2(x+y)} = \frac{x+y}{xy} \Leftrightarrow 9xy = 2x^2 + 4xy + 2y^2 \Leftrightarrow 2y^2 - 5xy + 2x^2 = 0$$

$y = \frac{5x \pm \sqrt{25x^2 - 16x^2}}{4} = \frac{5x \pm 3x}{4}$. It follows that all couples $(x, 2x)$ and $(x, \frac{1}{2}x)$ without $(0,0)$ are solutions of the first equation. We are putting these solutions in the second equation. From this equation we have $x^2 + y^2 = 5$, $|x| \ge \sqrt{2}$ and $y \le \sqrt{3}$. Further, putting $(x, 2x)$ the first condition becomes $5x^2 = 5 \Leftrightarrow x = \pm 1$. And we have solutions $(1,2)$ and $(-1,-2)$ but these couples don't satisfy the condition $|x| \ge \sqrt{2}$. Analogously putting $(x, \frac{1}{2}x)$ the first condition becomes $x^2 + \frac{1}{4}x^2 = 5 \Leftrightarrow x^2 = 4$. Here we have solutions $(2,1)$ and $(-2,-1)$ which satisfy our conditions.