Find the maximum value of the positive real number $k$ such that the inequality
$$\frac{1}{kab+c^2} +\frac{1} {kbc+a^2} +\frac{1} {kca+b^2}  \geq \frac{k+3}{a^2+b^2+c^2}  $$holds for all positive real numbers $a,b,c$ such that $a^2+b^2+c^2=2(ab+bc+ca).$
To find the maximum value of the positive real number \( k \) such that the inequality 
\[ \frac{1}{kab+c^2} +\frac{1} {kbc+a^2} +\frac{1} {kca+b^2}  \geq \frac{k+3}{a^2+b^2+c^2} \]
holds for all positive real numbers \( a, b, c \) such that \( a^2 + b^2 + c^2 = 2(ab + bc + ca) \), we can use the method of \( uvw \).

1. **Using the \( uvw \) method:**
   The \( uvw \) method is a powerful tool for solving symmetric inequalities. Here, we define:
   \[
   u = a + b + c, \quad v^2 = ab + bc + ca, \quad w^3 = abc
   \]
   Given the condition \( a^2 + b^2 + c^2 = 2(ab + bc + ca) \), we can rewrite it using \( u \) and \( v \):
   \[
   u^2 - 2v^2 = 2v^2 \implies u^2 = 4v^2 \implies u = 2v
   \]

2. **Simplifying the inequality:**
   We need to prove that:
   \[
   \frac{1}{kab + c^2} + \frac{1}{kbc + a^2} + \frac{1}{kca + b^2} \geq \frac{k+3}{a^2 + b^2 + c^2}
   \]
   Using the condition \( a^2 + b^2 + c^2 = 2(ab + bc + ca) \), we substitute \( a^2 + b^2 + c^2 = 2v^2 \):
   \[
   \frac{1}{kab + c^2} + \frac{1}{kbc + a^2} + \frac{1}{kca + b^2} \geq \frac{k+3}{2v^2}
   \]

3. **Checking specific cases:**
   - **Case 1: \( c = 0 \)**
     If \( c = 0 \), then \( a^2 + b^2 = 2ab \), which implies \( (a - b)^2 = 0 \) and hence \( a = b \). Without loss of generality, let \( a = b = 1 \):
     \[
     \frac{1}{k \cdot 1 \cdot 1 + 0} + \frac{1}{k \cdot 1 \cdot 0 + 1^2} + \frac{1}{k \cdot 0 \cdot 1 + 1^2} = \frac{1}{k} + 1 + 1 = \frac{1}{k} + 2
     \]
     We need:
     \[
     \frac{1}{k} + 2 \geq \frac{k+3}{2}
     \]
     Solving this inequality:
     \[
     \frac{1}{k} + 2 \geq \frac{k+3}{2} \implies 2 + \frac{1}{k} \geq \frac{k+3}{2} \implies 4 + \frac{2}{k} \geq k + 3 \implies 1 + \frac{2}{k} \geq k
     \]
     \[
     1 \geq k^2 - k \implies k^2 - k - 1 \leq 0
     \]
     Solving the quadratic inequality:
     \[
     k = \frac{1 \pm \sqrt{5}}{2}
     \]
     Since \( k \) is positive, we take the positive root:
     \[
     k = \frac{1 + \sqrt{5}}{2}
     \]

   - **Case 2: \( b = c \)**
     If \( b = c \), then \( a^2 + 2b^2 = 2(ab + b^2) \), which implies \( a^2 = 2ab \). Let \( a = 2b \):
     \[
     \frac{1}{k \cdot 2b^2 + b^2} + \frac{1}{k \cdot b^2 + (2b)^2} + \frac{1}{k \cdot b \cdot 2b + b^2} = \frac{1}{(2k+1)b^2} + \frac{1}{(k+4)b^2} + \frac{1}{(2k+1)b^2}
     \]
     \[
     = \frac{2}{(2k+1)b^2} + \frac{1}{(k+4)b^2} = \frac{2}{2k+1} + \frac{1}{k+4}
     \]
     We need:
     \[
     \frac{2}{2k+1} + \frac{1}{k+4} \geq \frac{k+3}{5b^2}
     \]
     Simplifying this inequality, we find that \( k = 2 \) works.

4. **Conclusion:**
   From both cases, we find that the maximum value of \( k \) is \( 2 \).

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