Hard system of equations for real numbers

Question

$\begin{cases} x+\frac{3x-y}{x^2+y^2}=3 \\ y-\frac{x+3y}{x^2+y^2}=0 \end{cases}$

I supposed that $\frac{1}{x^2+y^2}=u$. Then I develop and find the equation $(u+1)(x-y)+3u(x+y)=3$. But I got stuck. Anybody have an idea?

Answer 1

Solution via complex numbers - https://brilliant.org/problems/most-beautiful-problem/

Answer 2

Hint:Write $$\frac{x+3y}{3-x}=x^2+y^2$$ and $$\frac{x+3y}{y}=(x^2+y^2)$$ so $$\frac{3x-y}{3-x}=\frac{x+3y}{y}$$ From here we get $$3x+9y-6xy=x^2-y^2$$ and you can eliminate one variable. $$x=\frac{1}{2}(3-6y\pm\sqrt{9+40y^2})$$