How to solve a system of equations? How can I solve this system of equations?

$$ y^2 = x^3 - 3x^2 + 2x$$
$$ x^2 = y^3 - 3y^2 + 2y$$


$$y^2 = x^3 - 3x^2 + 2x$$
$$x^2 = y^3 - 3y^2 + 2y$$
Subtracring $(1.)$ from $(2.)$ , 
$$\begin{align}  x^2 - y^2 & = y^3 - x^3 + 3x^2 - 3y^2 + 2y-2x 
 \\  \color{#f04}{(x+y)(x-y)}& \color{#f04}{= (y-x)(y^2+x^2+xy) +3(x+y)(x-y) + 2(y-x)} \\
  \end{align}$$
If $(x-y)=0 \implies x=y$ , Which gives $\color{#39f}{x^2 = x^3 - 3x^2 + 2x}$
, Which yields $\color{#4b9}{x =0 , 2\pm \sqrt 2}$
If $x \ne y$ , Then :
$$x+y = -(x^2 + y^2 +xy) + 3(x+y) - 2$$
$$(x^2+y^2+xy) = 2(x+y-1)$$
$$x^2 + (y-2)x + (y^2-2y+2) = 0$$
Using Quadratic formula , we have obtain the discriminant :
$$D = (y-2)^2 - 4(y^2-2y+2)$$
$$D = -3y^2 + 4y - 4 \implies \boxed{D = -3 \left(y - \frac23 \right)^2 - \frac83}$$
We can observe that $\color{#da0}{D\lt 0}$ for all values of $y$. Hence the equation has no solutions for real $x,y$.
Hence the only solutions are $(x , y) = (0,0) \,\, , (2\pm\sqrt2,2\pm\sqrt2)$
