Determine all possible values of the expression $xy+yz+zx$ with real numbers $x, y, z$ satisfying the conditions $x^2-yz = y^2-zx = z^2-xy = 2$.
1. Given the conditions:
   \[
   x^2 - yz = 2, \quad y^2 - zx = 2, \quad z^2 - xy = 2
   \]
   we start by analyzing the equations pairwise.

2. Consider the first two equations:
   \[
   x^2 - yz = y^2 - zx
   \]
   Rearrange to get:
   \[
   x^2 - y^2 = yz - zx
   \]
   Factor both sides:
   \[
   (x - y)(x + y) = z(y - x)
   \]
   Simplify:
   \[
   (x - y)(x + y + z) = 0
   \]
   This implies either \(x = y\) or \(x + y + z = 0\).

3. Similarly, consider the first and third equations:
   \[
   x^2 - yz = z^2 - xy
   \]
   Rearrange to get:
   \[
   x^2 - z^2 = yz - xy
   \]
   Factor both sides:
   \[
   (x - z)(x + z) = y(z - x)
   \]
   Simplify:
   \[
   (x - z)(x + y + z) = 0
   \]
   This implies either \(x = z\) or \(x + y + z = 0\).

4. Finally, consider the second and third equations:
   \[
   y^2 - zx = z^2 - xy
   \]
   Rearrange to get:
   \[
   y^2 - z^2 = zx - xy
   \]
   Factor both sides:
   \[
   (y - z)(y + z) = x(z - y)
   \]
   Simplify:
   \[
   (y - z)(x + y + z) = 0
   \]
   This implies either \(y = z\) or \(x + y + z = 0\).

5. From the above, we see that the common solution is \(x + y + z = 0\) since \(x = y\), \(x = z\), or \(y = z\) would contradict the given conditions \(x^2 - yz = 2\), \(y^2 - zx = 2\), and \(z^2 - xy = 2\).

6. Now, add the three given equations:
   \[
   x^2 - yz + y^2 - zx + z^2 - xy = 6
   \]
   Group the terms:
   \[
   x^2 + y^2 + z^2 - (xy + yz + zx) = 6
   \]
   Since \(x + y + z = 0\), we have:
   \[
   (x + y + z)^2 = 0 \implies x^2 + y^2 + z^2 + 2(xy + yz + zx) = 0
   \]
   Substitute this into the previous equation:
   \[
   0 + 2(xy + yz + zx) - (xy + yz + zx) = 6
   \]
   Simplify:
   \[
   x^2 + y^2 + z^2 - (xy + yz + zx) = 6
   \]
   Substitute \(x^2 + y^2 + z^2 = -2(xy + yz + zx)\):
   \[
   -2(xy + yz + zx) - (xy + yz + zx) = 6
   \]
   Simplify:
   \[
   -3(xy + yz + zx) = 6
   \]
   Solve for \(xy + yz + zx\):
   \[
   xy + yz + zx = -2
   \]

The final answer is \(\boxed{-2}\).