Find the maximum value of $M =\frac{x}{2x + y} +\frac{y}{2y + z}+\frac{z}{2z + x}$ , $x,y, z > 0$
To find the maximum value of \( M = \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \) for \( x, y, z > 0 \), we will use the method of inequalities.

1. **Initial Setup:**
   We need to show that:
   \[
   \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq 1
   \]

2. **Using the AM-GM Inequality:**
   Consider the following application of the AM-GM (Arithmetic Mean-Geometric Mean) inequality:
   \[
   \frac{x}{2x + y} \leq \frac{x}{2x} = \frac{1}{2}
   \]
   Similarly,
   \[
   \frac{y}{2y + z} \leq \frac{y}{2y} = \frac{1}{2}
   \]
   and
   \[
   \frac{z}{2z + x} \leq \frac{z}{2z} = \frac{1}{2}
   \]

3. **Summing the Inequalities:**
   Adding these inequalities, we get:
   \[
   \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}
   \]
   However, this approach does not directly help us to prove the desired inequality. We need a different approach.

4. **Using the Titu's Lemma (Cauchy-Schwarz in Engel Form):**
   Titu's Lemma states that for non-negative real numbers \(a_1, a_2, \ldots, a_n\) and \(b_1, b_2, \ldots, b_n\),
   \[
   \frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \cdots + \frac{a_n^2}{b_n} \geq \frac{(a_1 + a_2 + \cdots + a_n)^2}{b_1 + b_2 + \cdots + b_n}
   \]
   Applying Titu's Lemma to our problem, we set \(a_1 = \sqrt{x}\), \(a_2 = \sqrt{y}\), \(a_3 = \sqrt{z}\), \(b_1 = 2x + y\), \(b_2 = 2y + z\), \(b_3 = 2z + x\):
   \[
   \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \geq \frac{(x + y + z)^2}{2x + y + 2y + z + 2z + x} = \frac{(x + y + z)^2}{3(x + y + z)} = \frac{x + y + z}{3}
   \]

5. **Conclusion:**
   Since \( \frac{x + y + z}{3} \leq 1 \) for all positive \(x, y, z\), we have:
   \[
   \frac{x}{2x + y} + \frac{y}{2y + z} + \frac{z}{2z + x} \leq 1
   \]
   Therefore, the maximum value of \( M \) is \(1\).

The final answer is \(\boxed{1}\).