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?

$$\displaystyle x+\frac{3x-y}{x^2+y^2}=3\cdots (1)$$
$$\displaystyle y-\frac{x+3y}{x^2+y^2}=0\cdots\cdots(2)\times i  $$
Now adding these two equations 
and substituting $z=x+iy$ and $\bar{z}=x-iy$
and $$|z|^2=x^2+y^2$$
So we have $$z+\frac{3-i}{z}=3\Rightarrow z^2-3z+(3-i)=0$$
On solving that equation we have 
$$z=2+i\;\;,1-i$$
So we get $$(x,y)=(2,1)\;\;,(1,-1)$$
