The inequality is equivalent to
$$\frac{x^2}{x+yz} + \frac{y^2}{y+zx} + \frac{z^2}{z+xy} \ge M. \quad (1)$$
Since $x, y, z > 0$, from Cauchy-Schwarz inequality we find
$$\begin{aligned} & \left( \frac{x^2}{x+yz} + \frac{y^2}{y+zx} + \frac{z^2}{z+xy} \right) (x+yz+y+zx+z+xy) \ge (x+y+z)^2 \\ \Leftrightarrow & \frac{x^2}{x+yz} + \frac{y^2}{y+zx} + \frac{z^2}{z+xy} \ge \frac{(x+y+z)^2}{x+y+z+xy+yz+zx} \\ \Leftrightarrow & \frac{x^2}{x+yz} + \frac{y^2}{y+zx} + \frac{z^2}{z+xy} \ge \frac{(x+y+z)^2}{x+y+z+1}, \end{aligned} \quad (2)$$
From $(x+y+z)^2 \ge 3(xy+yz+zx)$, it follows that: $x+y+z \ge \sqrt{3}$, where the equality holds when $x=y=z=\sqrt{3}/3$.
Now we consider the function $f(u) = \frac{u^2}{u+1}$, $u \ge \sqrt{3}$, which is strictly increasing, because for $u > v \ge \sqrt{3}$ we have $f(u) > f(v)$. In fact we have
$$\begin{align*} f(u) > f(v) &\Leftrightarrow \frac{u^2}{u+1} > \frac{v^2}{v+1} \\ &\Leftrightarrow u^2v + u^2 - uv^2 - v^2 > 0 \\ &\Leftrightarrow (u-v)(uv+u+v) > 0 \\ &\Leftrightarrow u-v > 0, \text{ since } uv+u+v > 0. \end{align*}$$
Therefore $f(u) \ge f(\sqrt{3}) = \frac{3}{\sqrt{3}+1}$, and hence from inequality (2) we have:
$$\frac{x^2}{x+yz} + \frac{y^2}{y+zx} + \frac{z^2}{z+xy} \ge \frac{(x+y+z)^2}{x+y+z+1} \ge \frac{3}{\sqrt{3}+1},$$
for all $x, y, z > 0$ with $xy + yz + zx = 1$. Hence: $M = \frac{3}{\sqrt{3}+1}$.