Find  the largest constant $k$ such that the inequality
$$a^2+b^2+c^2-ab-bc-ca \ge k \left|\frac{a^3-b^3}{a+b}+\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}\right|$$
holds for any for non negative real numbers $a,b,c$ with  $(a+b)(b+c)(c+a)>0$.
To find the largest constant \( k \) such that the inequality 
\[ a^2 + b^2 + c^2 - ab - bc - ca \ge k \left| \frac{a^3 - b^3}{a + b} + \frac{b^3 - c^3}{b + c} + \frac{c^3 - a^3}{c + a} \right| \]
holds for any non-negative real numbers \( a, b, c \) with \((a+b)(b+c)(c+a) > 0\), we proceed as follows:

1. **Simplify the Right-Hand Side:**
   We start by simplifying the expression inside the absolute value on the right-hand side. Notice that:
   \[
   \frac{a^3 - b^3}{a + b} = a^2 - ab + b^2
   \]
   This follows from the factorization of the difference of cubes:
   \[
   a^3 - b^3 = (a - b)(a^2 + ab + b^2)
   \]
   Therefore,
   \[
   \frac{a^3 - b^3}{a + b} = a^2 - ab + b^2
   \]
   Similarly,
   \[
   \frac{b^3 - c^3}{b + c} = b^2 - bc + c^2
   \]
   and
   \[
   \frac{c^3 - a^3}{c + a} = c^2 - ca + a^2
   \]

2. **Combine the Simplified Terms:**
   Adding these expressions together, we get:
   \[
   \frac{a^3 - b^3}{a + b} + \frac{b^3 - c^3}{b + c} + \frac{c^3 - a^3}{c + a} = (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2)
   \]
   Simplifying further, we have:
   \[
   = a^2 + b^2 + c^2 - ab - bc - ca + a^2 + b^2 + c^2
   \]
   \[
   = 2(a^2 + b^2 + c^2 - ab - bc - ca)
   \]

3. **Substitute Back into the Inequality:**
   The inequality now becomes:
   \[
   a^2 + b^2 + c^2 - ab - bc - ca \ge k \left| 2(a^2 + b^2 + c^2 - ab - bc - ca) \right|
   \]
   Simplifying the absolute value, we get:
   \[
   a^2 + b^2 + c^2 - ab - bc - ca \ge 2k \left| a^2 + b^2 + c^2 - ab - bc - ca \right|
   \]

4. **Analyze the Inequality:**
   For the inequality to hold for all non-negative \( a, b, c \) with \((a+b)(b+c)(c+a) > 0\), we need:
   \[
   1 \ge 2k
   \]
   Therefore,
   \[
   k \le \frac{1}{2}
   \]

5. **Verify the Largest \( k \):**
   To verify that \( k = \frac{1}{2} \) is indeed the largest possible value, we substitute \( k = \frac{1}{2} \) back into the inequality:
   \[
   a^2 + b^2 + c^2 - ab - bc - ca \ge \frac{1}{2} \left| 2(a^2 + b^2 + c^2 - ab - bc - ca) \right|
   \]
   Simplifying, we get:
   \[
   a^2 + b^2 + c^2 - ab - bc - ca \ge a^2 + b^2 + c^2 - ab - bc - ca
   \]
   This is always true, confirming that \( k = \frac{1}{2} \) is the largest possible value.

The final answer is \( \boxed{ k = \frac{1}{2} } \).