3.9. $\left\{\begin{array}{l}x+y+x y=2+3 \sqrt{2}, \\ x^{2}+y^{2}=6 .\end{array}\right.$
3.9. Let $u=x+y$ and $v=x y$. Then $u+v=2+3 \sqrt{2}$ and $u^{2}-2 v=6$, so $u^{2}+2 u=6+2(2+3 \sqrt{2})=10+6 \sqrt{2}$. Therefore, $u=-1 \pm \sqrt{11+6 \sqrt{2}}=-1 \pm(3+\sqrt{2})$, i.e., $u=2+\sqrt{2}$ or $-4-\sqrt{2}$. In this case, $v=2+3 \sqrt{2}-u=2 \sqrt{2}$ or $6+4 \sqrt{2}$. If $u=-4-\sqrt{2}$ and $v=6+4 \sqrt{2}$, then $(x-y)^{2}=(x+y)^{2}-4 x y=(4+\sqrt{2})^{2}-4(6+4 \sqrt{2})<0$. Therefore, $x+y=2+\sqrt{2}$ and $x y=2 \sqrt{2}$. This system of equations has two solutions: $(x, y)=(2, \sqrt{2})$ or $(\sqrt{2}, 2)$. Both are solutions to the original system of equations.