Solve the simultaneous equations $x + \frac{3x-y}{x^2+y^2} = 3 $, $ y – \frac{x+3y}{x^2+y^2} = 0$ Find all solution in $\mathbb{R}$ for the following system of equations:
\begin{cases} x + \frac{3x-y}{x^2+y^2} = 3 \\ y – \frac{x+3y}{x^2+y^2} = 0 \end{cases} 

I've tried few method, but none bring a success.
First I tried to substitute, and tried to express $y$ in terms of $x$, but I wasn't able to do that. I tried to use $y = \frac{x+3y}{x^2+y^2}$, but as you guess that substitution made the things even more complicated.
Next I tried to multiply the second equation with $\frac{3x-y}{x+3y}$, assuming it's not zero. And I end up with:
\begin{cases} x + \frac{3x-y}{x^2+y^2} = 3 \\ y\left(\frac{3x-y}{x+3y}\right) – \frac{x+3y}{x^2+y^2} = 0 \end{cases}
Now I add the together and I end up with:
$$x + y\left(\frac{3x-y}{x+3y}\right) = 3$$
But solution to this equation aren't solution to the system, because there are infinite amount of them.
Next I tried just to add them and I end up with:
$$x+y \frac{2x-3y}{x^2+y^2} = 3$$
But again I faced the same problem, this equation has an infinite amount of solution, unlike the system of equations.
So how can I solve this system of equations?

Complexify. Let $z = x+iy$. Then for $x^2+y^2 \neq 0$:
$$\begin{align}
&&\left(x + \frac{3x-y}{x^2+y^2}\right) + i\left(y - \frac{x+3y}{x^2+y^2}\right) &= (3+i0)\\
&\iff& (x+iy) + \frac{(3-i)(x-iy)}{x^2+y^2} &= 3\\
&\iff& z + \frac{3-i}{z} &= 3\\
&\iff& z^2 - 3z + (3-i) &= 0\\
&\iff& \left(z-\frac32\right)^2 &= i - \frac34\\
&\iff& z &= \frac{3 \pm \sqrt{4i-3}}{2}.
\end{align}$$
Now, $(1+2i)^2 = -3+4i$, so the solutions are
$$z = \frac{3 \pm (1+2i)}{2},$$
either $z = 2+i$ or $z = 1-i$, which, in real form, yields $(x,y) = (2,1)$ or $(x,y) = (1,-1)$.
