## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0 .\end{array}\right.$

Subtracting the second equation of the system from the first, we get

$$
-\frac{x}{y}+\frac{y}{x}=\frac{5}{6}
$$

Let $\frac{x}{y}=t$. Then we have $6 t^{2}+5 t-6=0$, from which $t_{1}=-\frac{3}{2}, t_{2}=\frac{2}{3}$. Thus,

$$
\text { we have } \frac{x}{y}=-\frac{3}{2} \text { or } \frac{x}{y}=\frac{2}{3} \text { and } x_{1}=-\frac{3}{2} y, x_{2}=\frac{2}{3} y .
$$

From the first equation of the system, we have: $-\frac{3}{2} y^{2}+\frac{3}{2}=\frac{16}{3}$, or $y^{2}=-\frac{23}{9}$ (no solutions), and $\frac{2}{3} y^{2}-\frac{2}{3}=\frac{16}{3}, y^{2}=9 \Rightarrow y_{1}=-3, y_{2}=3$. Then $x_{1}=-2, x_{2}=2$.

Answer: $(2 ; 3),(-2 ;-3)$.
6.211. $\left\{\begin{array}{l}x y-\frac{x}{y}=\frac{16}{3}, \\ x y-\frac{y}{x}=\frac{9}{2} .\end{array}\right.$