3. In the set of real numbers, solve the system of equations

$$
\left\{\begin{array}{l}
x+\frac{3 x-y}{x^{2}+y^{2}}=3 \\
y-\frac{x+3 y}{x^{2}+y^{2}}=0
\end{array}\right.
$$
Solution. We multiply the first equation by $y$, the second equation by $x$, and add the resulting equations, obtaining

$$
x y+\frac{3 x y-y^{2}}{x^{2}+y^{2}}+x y-\frac{x^{2}+3 x y}{x^{2}+y^{2}}=3 y, \text { i.e. } 2 x y-1=3 y
$$

Since $y \neq 0$, we get $x=\frac{3 y+1}{2 y}$. If we substitute this into the second equation, we sequentially obtain

$$
\begin{aligned}
& y-\frac{\frac{3 y+1}{2 y}+3 y}{\left(\frac{3 y+1}{2 y}\right)^{2}+y^{2}}=0 \\
& y-\frac{2 y\left(6 y^{2}+3 y+1\right)}{9 y^{2}+6 y+1+4 y^{4}}=0 \\
& 4 y^{4}+9 y^{2}+6 y+1-12 y^{2}-6 y-2=0 \\
& 4 y^{4}-3 y^{2}-1=0
\end{aligned}
$$

Let $t=y^{2}$. Then the solutions to the quadratic equation $4 t^{2}-3 t-1=0$ are $t_{1}=1$ and $t_{2}=-\frac{1}{4}$. Since the system must be solved in the set of real numbers, we get $y^{2}=1$, from which we find $y_{1}=1$ and $y_{2}=-1$. By substituting into $x=\frac{3 y+1}{2 y}$, we get $x_{1}=2, x_{2}=1$.

Finally, the solutions to the system are $(2,1)$ and $(1,-1)$.