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?

Make the substitution $x = r\cos(\theta)$ and $y = r\sin(\theta)$ in order to get
\begin{cases}
r^{2}\cos(\theta) + 3\cos(\theta) - \sin(\theta) = 3r\\
r^{2}\sin(\theta) - \cos(\theta) - 3\sin(\theta) = 0\\
\end{cases}
From the second equation, it results that $\cos(\theta) = (r^{2}-3)\sin(\theta)$. Therefore we get
\begin{align*}
(r^{2} + 3)\cos(\theta) - \sin(\theta) & = 3r \Longleftrightarrow (r^{2}+3)(r^{2}-3)\sin(\theta) - \sin(\theta) = 3r \Longleftrightarrow\\
 (r^{4}- 10)\sin(\theta) & = 3r \Longleftrightarrow\sin(\theta) = \frac{3r}{r^{4}-10}
\end{align*}
Consequently, we have
\begin{align*}
&\cos^{2}(\theta) + \sin^{2}(\theta) = 1 \Leftrightarrow\left[\frac{3r(r^{2}-3)}{r^{4}-10}\right]^{2} + \left[\frac{3r}{r^{4}-10}\right]^{2} = 1 \Leftrightarrow\\\\
& 9r^{2}(r^{4}-6r^{2}+9) + 9r^{2} = (r^{4}-10)^{2} \Leftrightarrow 9r^{6}-54r^{4} + 90r^{2} = r^{8} - 20r^{4} + 100 \Leftrightarrow\\\\
& r^{8} - 9r^{6} + 34r^{4} - 90r^{2} + 100 = 0 \Leftrightarrow (r^{8} - 5r^{6}) - (4r^{6} - 20r^{4})  + (14r^{4} - 70r^{2}) - (20r^{2} - 100) = 0 \Leftrightarrow\\\\
& r^{6}(r^{2}-5) - 4r^{4}(r^{2}-5) + 14r^{2}(r^{2}-5) - 20(r^{2}-5) = (r^{6} - 4r^{4} + 14r^{2} - 20)(r^{2}-5) = 0 \Leftrightarrow\\\\
& [(r^{6} - 2r^{4}) - (2r^{4} - 4r^{2}) + (10r^{2}-20)](r^{2}-5) = 0 \Leftrightarrow (r^{4}-2r^{2}+10)(r^{2}-2)(r^{2}-5) = 0 \Leftrightarrow\\\\
& (r^{2} - 2)(r^{2} - 5) = 0 \Leftrightarrow r\in\{\sqrt{2},\sqrt{5}\}\quad\text{since}\quad r\geq 0
\end{align*} 
Finally, for each value of $r$ there corresponds a solution. More precisely, the solution set is described by $$S = \{(1,-1),(2,1)\}$$
