For $x, y, z \in \mathbb{R}$, such that $xyz \neq 0$, the system is written:
$$x^3 y z = z^2 - 2y^2 \quad (1), \quad y^3 z x = x^2 - 2z^2 \quad (2), \quad z^3 x y = y^2 - 2x^2 \quad (3)$$
Using summation by parts we find:
$$xyz(x^2 + y^2 + z^2) = -(x^2 + y^2 + z^2) \Leftrightarrow (x^2 + y^2 + z^2)(xyz + 1) = 0.$$
Since $xyz \neq 0$ we have $x^2 + y^2 + z^2 > 0$, and so: $xyz = -1$ (4)
Using equation (4) in the system of (1)-(3) we get:
$$x^2 = -z^2 + 2y^2 \quad (5), \quad y^2 = -x^2 + 2z^2 \quad (6), \quad z^2 = -y^2 + 2x^2 \quad (7)$$
From (5) and (6) we get $y^2 = z^2$, while from (6) and (7) we get $x^2 = z^2$, and so:
$$x^2 = y^2 = z^2 \Leftrightarrow x = y = \pm z \quad \text{or} \quad x = -y = \pm z. \qquad (8)$$
Finally from equations (8) and (4) we have the solutions:
$$(x, y, z) = (-1, -1, -1), \quad (x, y, z) = (1, 1, -1), \quad (x, y, z) = (1, -1, 1), \quad (x, y, z) = (-1, 1, 1).$$