The first term $\frac{x(2x-y)}{y(2z+x)} + 1 = \frac{2(x^2+yz)}{y(2z+x)}$. The similar transformation of two other terms yields:
$$f(x, y, z) = \frac{x^2 + yz}{y(2z+x)} + \frac{y^2 + zx}{z(2x+y)} + \frac{z^2 + xy}{x(2y+z)} \ge 2$$
By Cauchy-Schwarz inequality for positive $x_1, \dots, x_n$
$$(x_1 + \dots + x_n)\left(\frac{a_1^2}{x_1} + \dots + \frac{a_n^2}{x_n}\right) \ge (a_1 + \dots + a_n)^2 \quad (1)$$
Therefore,
$$g(x, y, z) = \frac{x^2}{y(2z+x)} + \frac{y^2}{z(2x+y)} + \frac{z^2}{x(2y+z)} \ge \frac{(x+y+z)^2}{3(xy+yz+zx)}$$
$$h(x, y, z) = \frac{z}{2z+x} + \frac{x}{2x+y} + \frac{y}{2y+z} \ge \frac{(x+y+z)^2}{2(x^2+y^2+z^2)+xy+yz+zx}$$
Therefore,
$$\begin{align*} f &= g + h \ge (x+y+z)^2 \left( \frac{1}{3(xy+yz+zx)} + \frac{1}{2(x^2+y^2+z^2)+xy+yz+zx} \right) \\ &= \frac{2(x+y+z)^4}{3(xy+yz+zx)(2(x^2+y^2+z^2)+xy+yz+zx)} \\ &= \frac{2(x+y+z)^4}{3(xy+yz+zx)(2(x+y+z)^2 - 3(xy+yz+zx))} \\ &\ge \frac{2(x+y+z)^4}{\left( \frac{3(xy+yz+zx) + 2(x+y+z)^2 - 3(xy+yz+zx)}{2} \right)^2} = 2 \end{align*}$$
(in the last inequality we applied AM-GM inequality). Done.