Find all the solutions of the system
 $$\begin{cases}  y^2 = x^3 - 3x^2 + 2x \\ x^2 = y^3 - 3y^2 + 2y  \end{cases}$$
1. **Identify the trivial solution:**
   - Given the system of equations:
     \[
     \begin{cases}
     y^2 = x^3 - 3x^2 + 2x \\
     x^2 = y^3 - 3y^2 + 2y
     \end{cases}
     \]
   - If \( x = 0 \), then substituting \( x = 0 \) into the first equation gives \( y^2 = 0 \), so \( y = 0 \).
   - Similarly, if \( y = 0 \), then substituting \( y = 0 \) into the second equation gives \( x^2 = 0 \), so \( x = 0 \).
   - Therefore, \((x, y) = (0, 0)\) is a solution.

2. **Assume \( x \neq 0 \) and \( y \neq 0 \):**
   - Subtract the second equation from the first:
     \[
     y^2 - x^2 = (x^3 - 3x^2 + 2x) - (y^3 - 3y^2 + 2y)
     \]
   - Simplify the right-hand side:
     \[
     y^2 - x^2 = x^3 - y^3 - 3(x^2 - y^2) + 2(x - y)
     \]
   - Factor both sides:
     \[
     (y - x)(y + x) = (x - y)(x^2 + xy + y^2) - 3(x - y)(x + y) + 2(x - y)
     \]
   - Factor out \((x - y)\) from the right-hand side:
     \[
     (x - y)[(x^2 + xy + y^2) - 3(x + y) + 2]
     \]
   - Simplify the expression inside the brackets:
     \[
     (x - y)[(x^2 + xy + y^2) - 3(x + y) + 2] = (x - y)[(x - 1)^2 + (y - 1)^2 + xy]
     \]
   - Therefore, we have:
     \[
     (x - y)[(x - 1)^2 + (y - 1)^2 + xy] = 0
     \]
   - This implies either \( x = y \) or \((x - 1)^2 + (y - 1)^2 + xy = 0\).

3. **Case 1: \( x = y \):**
   - Substitute \( x = y \) into the first equation:
     \[
     x^2 = x^3 - 3x^2 + 2x
     \]
   - Simplify and solve for \( x \):
     \[
     x^3 - 4x^2 + 2x = 0
     \]
     \[
     x(x^2 - 4x + 2) = 0
     \]
   - Since \( x \neq 0 \), solve the quadratic equation:
     \[
     x^2 - 4x + 2 = 0
     \]
   - Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
     \[
     x = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}
     \]
   - Therefore, the solutions in this case are \( (2 + \sqrt{2}, 2 + \sqrt{2}) \) and \( (2 - \sqrt{2}, 2 - \sqrt{2}) \).

4. **Case 2: \((x - 1)^2 + (y - 1)^2 + xy = 0\):**
   - This equation implies that the sum of squares of real numbers plus a product is zero, which is only possible if each term is zero or negative.
   - However, since squares of real numbers are non-negative, the only way this equation holds is if \( xy < 0 \).
   - Assume \( x < 0 \) and \( y > 0 \). Let \( x = -z \) where \( z > 0 \):
     \[
     (-z)^3 - 3(-z)^2 + 2(-z) = y^2
     \]
     \[
     -z^3 - 3z^2 - 2z = y^2
     \]
   - The left-hand side is negative, while the right-hand side is positive, leading to a contradiction.
   - Therefore, there are no solutions in this case.

Conclusion:
The solutions to the system are \((0, 0)\), \((2 + \sqrt{2}, 2 + \sqrt{2})\), and \((2 - \sqrt{2}, 2 - \sqrt{2})\).

The final answer is \( \boxed{ (0, 0) } \), \((2 + \sqrt{2}, 2 + \sqrt{2})\), and \((2 - \sqrt{2}, 2 - \sqrt{2})\).