Solution:
We have
$$\begin{aligned} & \frac{x^{2}+y z}{x^{2}+y z+1}+\frac{y^{2}+z x}{y^{2}+z x+1}+\frac{z^{2}+x y}{z^{2}+x y+1} \leq 2 \Leftrightarrow \\ & \frac{x^{2}+y z+1}{x^{2}+y z+1}+\frac{y^{2}+z x+1}{y^{2}+z x+1}+\frac{z^{2}+x y+1}{z^{2}+x y+1} \leq 2+\frac{1}{x^{2}+y z+1}+\frac{1}{y^{2}+z x+1}+\frac{1}{z^{2}+x y+1} \Leftrightarrow \\ & 3 \leq 2+\frac{1}{x^{2}+y z+1}+\frac{1}{y^{2}+z x+1}+\frac{1}{z^{2}+x y+1} \Leftrightarrow \\ & 1 \leq \frac{1}{x^{2}+y z+1}+\frac{1}{y^{2}+z x+1}+\frac{1}{z^{2}+x y+1} \\ & \frac{1}{x^{2}+y z+1}+\frac{1}{y^{2}+z x+1}+\frac{1}{z^{2}+x y+1} \geq \frac{9}{x^{2}+y z+1+y^{2}+z x+1+z^{2}+x y+1}= \\ & \frac{9}{x^{2}+y^{2}+z^{2}+x y+y z+z x+3} \geq \frac{9}{2 x^{2}+y^{2}+z^{2}+3}=1 \end{aligned}$$
The first inequality: AM-GM inequality (also can be achieved with Cauchy-Bunjakowski-Schwarz inequality). The second inequality: $x y+y z+z x \leq x^{2}+y^{2}+z^{2}$ (there are more ways to prove it, AM-GM, full squares etc.)