3.12. $\left\{\begin{array}{l}x+\frac{3 x-y}{x^{2}+y^{2}}=3, \\ y-\frac{x+3 y}{x^{2}+y^{2}}=0 .\end{array}\right.$
3.12. Multiply the first equation by $y$, the second by $x$ and add the resulting equations. As a result, we get $2 x y - 1 = 3 y$. In particular, $y \neq 0$, so $x = \frac{3}{2} + \frac{1}{2 y}$. Substituting this expression into the second equation, after some transformations, we get $4 y^{4} - 3 y^{2} - 1 = 0$. Considering that $y^{2} \geqslant 0$, we obtain $y^{2} = 1$, i.e., $y_{1} = 1$ and $y_{2} = -1$. These values of $y$ correspond to $x_{1} = 2$ and $x_{2} = 1$.