We make a substitution: $a = \frac{2x}{2x+y}$, $b = \frac{2y}{2y+z}$, $c = \frac{2z}{2z+x}$, then the inequality will be as follows:
$$2 \cdot \left(\frac{x}{2x+y}\right)^2 + 2 \cdot \left(\frac{y}{2y+z}\right)^2 + 2 \cdot \left(\frac{z}{2z+x}\right)^2 + \frac{9xyz}{(2x+y)(2y+z)(2z+x)} = \frac{a^2+b^2+c^2}{2} + \frac{9}{8}abc \le 1, \text{ or}$$
$$a^2 + b^2 + c^2 + \frac{9}{4}abc \le 2.$$

Consider the expression:
$$\begin{aligned} \left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right) &= \left(\frac{2x+y}{2x}-1\right)\left(\frac{2y+z}{2y}-1\right)\left(\frac{2z+x}{2z}-1\right) = \frac{y}{2x} \cdot \frac{z}{2y} \cdot \frac{x}{2z} = \frac{1}{8}, \text{ or} \\ \frac{(1-a)}{a} \cdot \frac{(1-b)}{b} \cdot \frac{(1-c)}{c} &= \frac{1}{8} \Leftrightarrow 8(1-a)(1-b)(1-c) = abc \Leftrightarrow 8(1-a)(1-b)(1-c) = abc \\ &\Leftrightarrow 8-8(a+b+c)+8(ab+bc+ca)-8abc = abc \Rightarrow \\ &\frac{9}{4}abc = 2-2(a+b+c)+2(ab+bc+ca). \end{aligned}$$
Continue the proof using the obtained conditions:
$$\begin{aligned} & a^2 + b^2 + c^2 + \frac{9}{4}abc - 2 = \\ & = (a+b+c)^2 - 2(ab+bc+ca) + 2 - 2(a+b+c) + 2(ab+bc+ca) - 2 = \\ & = (a+b+c)^2 - 2(a+b+c) \le 0. \end{aligned}$$

The last inequality holds when $a+b+c \le 2$.

Perform transformations:
$$\begin{aligned} a+b+c &= \frac{2x}{2x+y} + \frac{2y}{2y+z} + \frac{2z}{2z+x} = 1 - \frac{y}{2x+y} + 1 - \frac{z}{2y+z} + 1 - \frac{x}{2z+x} = 2 + \left(1 - \frac{y}{2x+y} - \frac{z}{2y+z} - \frac{x}{2z+x}\right) \le 2 \\ &\Leftrightarrow \frac{y}{2x+y} + \frac{z}{2y+z} + \frac{x}{2z+x} \ge 1. \end{aligned}$$

Prove the last inequality using Cauchy–Schwarz inequality:
$$\left( \frac{y}{2x+y} + \frac{z}{2y+z} + \frac{x}{2z+x} \right) (x+y+z)^2 = \left( \frac{y}{2x+y} + \frac{z}{2y+z} + \frac{x}{2z+x} \right) (y(2x+y) + z(2y+z) + x(2z+x)) \ge$$
$$\geq \left( \sqrt{\frac{y}{2x+y}} \cdot \sqrt{y(2x+y)} + \sqrt{\frac{z}{2y+z}} \cdot \sqrt{z(2y+z)} + \sqrt{\frac{x}{2z+x}} \cdot \sqrt{x(2z+x)} \right)^2 = (x+y+z)^2,$$
Q.E.D.