6. Let's highlight the solution $(0,0)$, by dividing both sides of the first equation by $x y$, and the second by $x^{2} y^{2}$, we get:
$\left\{\begin{array}{c}\left(\frac{x}{y}+\frac{y}{x}\right)(x+y)=15 \\ \left(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}\right)\left(x^{2}+y^{2}\right)=85\end{array}\right.$

Let $x+y=a$ and $\frac{x}{y}+\frac{y}{x}=b$, then the first equation becomes

$a b=15$. Considering that $\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}=b^{2}-2$ and $x y=\frac{a^{2}}{2+b}$, the second equation

can be written as: $\left(b^{2}-2\right) a^{2} b=85(2+b)$. Substituting the expression for $a$ from the first equation, we get $\frac{3}{b}\left(b^{2}-2\right) 15=17(2+b) \Rightarrow 14 b^{2}-17 b-45=0$. Then $b_{1}=-\frac{9}{7}, b_{2}=\frac{5}{2}$. Therefore, $a_{1}=\frac{15}{b}=-\frac{15}{\frac{9}{7}}=-\frac{35}{3}, a_{2}=6$. We have two systems: 1) $\left\{\begin{array}{l}x+y=6 \\ \frac{x}{y}+\frac{y}{x}=\frac{5}{2}\end{array}\right.$ or 2) $\left\{\begin{array}{l}x+y=-\frac{35}{3} \\ \frac{x}{y}+\frac{y}{x}=-\frac{9}{7}\end{array}\right.$. Solving 1) $\left\{\begin{array}{c}x+y=6 \\ x^{2}+y^{2}=\frac{5}{2} x y\end{array} \Rightarrow\left\{\begin{array}{c}x+y=6 \\ (x+y)^{2}=4.5 x y\end{array} \Rightarrow\left\{\begin{array}{c}x+y=6 \\ 4.5 x y=36\end{array} \Rightarrow\right.\right.\right.$ $\left\{\begin{array}{c}x+y=6 \\ x y=8\end{array}\right.$. Then, by the inverse Vieta's theorem, we have: $x=2 \quad y=4, x=4 \quad y=2$.

Solving the second system. $\left\{\begin{array}{l}x+y=-\frac{35}{3} \\ \frac{x}{y}+\frac{y}{x}+\frac{9}{7}=0\end{array}\left(\frac{y}{x}\right)^{2}+\frac{9}{7}\left(\frac{y}{x}\right)+1=0 \quad D<0\right.$ no solutions.
Answer: $(0,0),(2,4),(4,2)$.