Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$ 
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
  Solve in the set of real numbers.

The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^2+2)x+y^3-3y^2-4y=0.$$
But I have no idea how to solve this problem. How can I solve it?

Note that if $y=0$ then we get $-\frac 1x=0$ on second equation, so the system has no solutions.
So we can assume $y\neq 0$, set $\ x=ty\ $ and substitute.
We get $\begin{cases}t^3y^2+ty^2+3t-1=3t^2y+3y \\t^2y^2+y^2-t-3=0\end{cases}$
And this allow us to isolate $y^2=\dfrac{t+3}{t^2+1}$
Substituting in first equation rewritten $ty^2(t^2+1)+3t-1=3y(t^2+1)$ 
gives $t(t+3)+3t-1=3y(t^2+1)\iff 3y=\dfrac{t^2+6t-1}{t^2+1}$
Now be identifying $9y^2$ we get $(t^2+6t-1)^2=9(t+3)(t^2+1)$
We can remark $t=-1$ is a root $6^2=9(2)(2)$ else just graph it, and notice $-1$ and $2$ seems to be roots (so try to factor these terms).
Indeed the factorization is $$(t+1)(t-2)(t^2+4t+13)=0$$
Reporting in initial equations gives us the solutions:


*

*$t=2$ and $x=2y$ leads to $\begin{cases}2y+\frac 1y=3\\y=\frac 1y\end{cases}\iff \begin{cases}x=2\\y=1\end{cases}$

*$t=-1$ and $x=-y$ leads to $\begin{cases}-y-\frac 2y=3\\y=\frac 1y\end{cases}\iff \begin{cases}x=1\\y=-1\end{cases}$
