Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
{y^2} = {x^3} - 3{x^2} + 2x\\
{x^2} = {y^3} - 3{y^2} + 2y
\end{array} \right.\\
\Rightarrow {y^2} - {x^2} = {x^3} - {y^3} - 3{x^2} + 3{y^2} + 2x - 2y\\
\Rightarrow \left( {x + y} \right)\left( {y - x} \right) = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right) - \\
3\left( {x - y} \right)\left( {x + y} \right) + 2\left( {x - y} \right)\\
\Rightarrow \left( {x - y} \right)\left( {{x^2} + xy + {y^2} - 3x - 3y + 2 + x + y} \right) = 0\\
\Rightarrow \left( {x - y} \right)\left( {{x^2} + xy + {y^2} - 2x - 2y + 2} \right) = 0\\
\Rightarrow x = y\\
\Rightarrow {y^2} = {y^3} - 3{y^2} + 2y\\
\Rightarrow {y^3} - 4{y^2} + 2y = 0\\
\Rightarrow \left[ \begin{array}{l}
y = 0 = x\\
{y^2} - 4y + 2 = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = y = 0\\
x = y = 2 \pm \sqrt 2
\end{array} \right.
\end{array}$
