Solution:
For $x = y = z = 0$ the equality is valid.
Since $(x + y + z)^{2} \geq 0$ it is enough to prove that
$$\frac{x^{2}-y^{2}}{2 x^{2}+1}+\frac{y^{2}-z^{2}}{2 y^{2}+1}+\frac{z^{2}-x^{2}}{2 z^{2}+1} \leq 0$$
which is equivalent to the inequality
$$\frac{x^{2}-y^{2}}{x^{2}+\frac{1}{2}}+\frac{y^{2}-z^{2}}{y^{2}+\frac{1}{2}}+\frac{z^{2}-x^{2}}{z^{2}+\frac{1}{2}} \leq 0$$
Denote
$$a = x^{2} + \frac{1}{2}, \quad b = y^{2} + \frac{1}{2}, \quad c = z^{2} + \frac{1}{2}$$
Then (1) is equivalent to
$$\frac{a-b}{a}+\frac{b-c}{b}+\frac{c-a}{c} \leq 0$$
From the very well known $AG$ inequality it follows that
$$a^{2}b + b^{2}c + c^{2}a \geq 3abc$$
From the equivalencies
$$a^{2}b + b^{2}c + c^{2}a \geq 3abc \Leftrightarrow \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \geq -3 \Leftrightarrow \frac{a-b}{a} + \frac{b-c}{b} + \frac{c-a}{c} \leq 0$$
it follows that the inequality (2) is valid for positive real numbers $a, b, c$.