Let $x,y,z$ be real numbers such that the numbers $$\frac{1}{|x^2+2yz|},\quad\frac{1}{|y^2+2zx|},\quad\frac{1}{|z^2+2xy|}$$ are lengths of sides of a (non-degenerate) triangle. Determine all possible values of $xy+yz+zx$.
To determine all possible values of \( xy + yz + zx \) given that the numbers 
\[ \frac{1}{|x^2 + 2yz|}, \quad \frac{1}{|y^2 + 2zx|}, \quad \frac{1}{|z^2 + 2xy|} \]
are the lengths of the sides of a non-degenerate triangle, we need to analyze the conditions under which these expressions can form a triangle.

1. **Triangle Inequality Conditions**:
   For three numbers \(a, b, c\) to be the sides of a non-degenerate triangle, they must satisfy the triangle inequalities:
   \[
   a + b > c, \quad b + c > a, \quad c + a > b
   \]
   In our case, we have:
   \[
   \frac{1}{|x^2 + 2yz|} + \frac{1}{|y^2 + 2zx|} > \frac{1}{|z^2 + 2xy|}, \quad \frac{1}{|y^2 + 2zx|} + \frac{1}{|z^2 + 2xy|} > \frac{1}{|x^2 + 2yz|}, \quad \frac{1}{|z^2 + 2xy|} + \frac{1}{|x^2 + 2yz|} > \frac{1}{|y^2 + 2zx|}
   \]

2. **Special Cases**:
   Consider the cases where \(x = y = z\) and \(x = y = -2z\):
   - If \(x = y = z = t\), then:
     \[
     x^2 + 2yz = t^2 + 2t^2 = 3t^2
     \]
     Thus, the side lengths become:
     \[
     \frac{1}{|3t^2|}, \quad \frac{1}{|3t^2|}, \quad \frac{1}{|3t^2|}
     \]
     These are equal and can form a non-degenerate triangle if \(t \neq 0\).

   - If \(x = y = t\) and \(z = -2t\), then:
     \[
     x^2 + 2yz = t^2 + 2(t)(-2t) = t^2 - 4t^2 = -3t^2
     \]
     Thus, the side lengths become:
     \[
     \frac{1}{|-3t^2|}, \quad \frac{1}{|-3t^2|}, \quad \frac{1}{|-3t^2|}
     \]
     These are also equal and can form a non-degenerate triangle if \(t \neq 0\).

3. **General Case**:
   To show that \(xy + yz + zx\) cannot be zero, suppose to the contrary that \(xy + yz + zx = 0\). Then:
   \[
   \sum_{cyc} \frac{1}{|x^2 + 2yz|} = \sum_{cyc} \frac{1}{|(x-y)(x-z)|} = 0
   \]
   This implies that at least one of the terms must be zero, which contradicts the non-degenerate triangle condition.

4. **Conclusion**:
   From the special cases and the general analysis, we see that \(xy + yz + zx\) can take any real value except zero.

The final answer is \( \boxed{ xy + yz + zx } \) can be any real number except zero.