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?

If $x+3y=0$ so from the second equation we obtain $y=0$, which gives $x=0$, which is impossible. 
Thus, $x+3y\neq0$ and since $$\frac{(x^2+y^2)y}{x+3y}=1,$$ from the first equation we obtain:
$$x+\frac{3x-y}{x^2+y^2}\cdot\frac{(x^2+y^2)y}{x+3y}=3\sqrt{\frac{(x^2+y^2)y}{x+3y}}$$ or
$$x^2+6xy-y^2=3\sqrt{y(x^2+y^2)(x+3y)}$$ or with $x^2+6xy-y^2\geq0$
$$(x^2+6xy-y^2)^2=9y(x^2+y^2)(x+3y)$$ or
$$(x+y)(x-2y)(x^2+4xy+9y^2)=0,$$ which gives $y=-x$ or $x=2y$.
Can you end it now?
