Solve the system of equation
$$x+y+z=2;$$$$(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)=1;$$$$x^2(y+z)+y^2(z+x)+z^2(x+y)=-6.$$
1. **Given System of Equations:**
   \[
   \begin{cases}
   x + y + z = 2 \\
   (x+y)(y+z) + (y+z)(z+x) + (z+x)(x+y) = 1 \\
   x^2(y+z) + y^2(z+x) + z^2(x+y) = -6
   \end{cases}
   \]

2. **First Equation:**
   \[
   x + y + z = 2
   \]

3. **Second Equation:**
   \[
   (x+y)(y+z) + (y+z)(z+x) + (z+x)(x+y) = 1
   \]
   Expanding each term:
   \[
   (x+y)(y+z) = xy + yz + y^2 + yz
   \]
   \[
   (y+z)(z+x) = yz + zx + z^2 + zx
   \]
   \[
   (z+x)(x+y) = zx + xy + x^2 + xy
   \]
   Summing these:
   \[
   (x+y)(y+z) + (y+z)(z+x) + (z+x)(x+y) = xy + yz + y^2 + yz + yz + zx + z^2 + zx + zx + xy + x^2 + xy
   \]
   Simplifying:
   \[
   = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx
   \]
   Using the identity \((x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)\):
   \[
   (x+y+z)^2 + xy + yz + zx = 4 + xy + yz + zx = 1
   \]
   Therefore:
   \[
   4 + xy + yz + zx = 1 \implies xy + yz + zx = -3
   \]

4. **Third Equation:**
   \[
   x^2(y+z) + y^2(z+x) + z^2(x+y) = -6
   \]
   Using \(x + y + z = 2\):
   \[
   x^2(y+z) + y^2(z+x) + z^2(x+y) = x^2(2-x) + y^2(2-y) + z^2(2-z)
   \]
   Simplifying:
   \[
   = 2x^2 - x^3 + 2y^2 - y^3 + 2z^2 - z^3
   \]
   Grouping terms:
   \[
   = 2(x^2 + y^2 + z^2) - (x^3 + y^3 + z^3)
   \]
   Given:
   \[
   2(x^2 + y^2 + z^2) - (x^3 + y^3 + z^3) = -6
   \]
   Using the identity \(x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)\):
   \[
   x^3 + y^3 + z^3 - 3xyz = 2(4 + 3) = 14
   \]
   Therefore:
   \[
   x^3 + y^3 + z^3 = 14 + 3xyz
   \]
   Substituting back:
   \[
   2(x^2 + y^2 + z^2) - (14 + 3xyz) = -6
   \]
   Simplifying:
   \[
   2(x^2 + y^2 + z^2) - 14 - 3xyz = -6 \implies 2(x^2 + y^2 + z^2) - 3xyz = 8
   \]
   Using \(x^2 + y^2 + z^2 = 4 + 6 = 10\):
   \[
   2(10) - 3xyz = 8 \implies 20 - 3xyz = 8 \implies 3xyz = 12 \implies xyz = 0
   \]

5. **Finding Roots:**
   Since \(xyz = 0\), one of \(x, y, z\) must be zero. Assume \(x = 0\):
   \[
   y + z = 2 \quad \text{and} \quad yz = -3
   \]
   Solving the quadratic equation:
   \[
   t^2 - 2t - 3 = 0
   \]
   Factoring:
   \[
   (t-3)(t+1) = 0 \implies t = 3 \quad \text{or} \quad t = -1
   \]
   Therefore, the solutions are:
   \[
   (x, y, z) = (0, 3, -1), (0, -1, 3)
   \]
   Similarly, permuting \(x, y, z\) gives:
   \[
   (3, 0, -1), (3, -1, 0), (-1, 0, 3), (-1, 3, 0)
   \]

The final answer is \(\boxed{\{(0, 3, -1), (0, -1, 3), (3, 0, -1), (3, -1, 0), (-1, 0, 3), (-1, 3, 0)\}}\)