Solve the System of Equations in $x$ and $y$ \begin{equation}
x+\frac{3\,x-y}{x^2+y^2}=3 \tag{1}
\end{equation}
\begin{equation}
y=\frac{x+3\,y}{x^2+y^2} \tag{2}
\end{equation}
 A: I am assuming that you are solving this equation over the reals only. This approach is motivated because you have $x^2 + y^2$, and 'anti-symmetric' coefficients.
Set $z = x+iy $.
Take the first equation, add it to $i$ times the second equation, we get
$$ (x+iy) + \frac{ (3-i) ( x-iy) } { x^2 + y^2} = 3.$$
Converting this to $z$, we get
$$ z + \frac{ (3-i) \bar{z} } { |z|^2 } = 3.$$
Recall that $ z \bar{z}  = |z|^2$. Multiplying throughout by $z$, we get
$$ z^2 - 3z + (3-i) = 0.$$
This has solutions $ z= 1-i, z = 2+i.$ This yields $(x,y) = (1,-1), (2,1) $.

If you want more problems along the line of using complex substitution, try Advanced System Of Equations and Making An Equation Complex Doesn't Make It Harder.
A: Hint
Just by inspection, the quartic equation you wrote has two simple solutions $x=1$ and $x=2$. For the other roots, perform the polynomial division.
I am sure that you can take from here.
A: We have:
\begin{equation}
x\left(1+\frac{3}{x^2+y^2}\right)=3+\frac{y}{x^2+y^2} \tag{3}
\end{equation}
\begin{equation}
y\left(1-\frac{3}{x^2+y^2}\right)=\frac{x}{x^2+y^2} \tag{4}
\end{equation}
Multiplying Eqn $3$ with $y$ and Eqn $4$ with $x$ we get
\begin{equation}
x\,y\left(1+\frac{3}{x^2+y^2}\right)=3\,y+\frac{y^2}{x^2+y^2} \tag{5}
\end{equation}
\begin{equation}
x\,y\left(1-\frac{3}{x^2+y^2}\right)=\frac{x^2}{x^2+y^2} \tag{6}
\end{equation} Adding Eqns $5$ and $6$ we get
$$y=\frac{1}{2\,x-3}$$ Using $y=\frac{1}{2\,x-3}$ in Eqn $2$ we get Bi Quadratic in $x$ as:
$$4x^4-24x^3+57x^2-63x+26=0$$
Can i get a Hint without involvement of BiQuadratic...
