From the form of the equations it is immediate that $xyz \neq 0$. Two of the numbers $x, y, z$ have to be of the same sign; then the right-hand side of the equation where the ratio of these two numbers occurs is positive, hence so must be the corresponding left-hand side, which implies that the third of the numbers $x, y, z$ must also have the same sign as the first and the second. Thus either $x, y, z > 0$, or $x, y, z < 0$. Let us consider only the former case (the latter can be reduced to it by passing from the solution $(x, y, z)$ to the solution $(-x, -y, -z)$).

Multiply the first two equations of the system by the expression $xyz$ and then subtract them; this gives, upon a small manipulation, $z - x = y(x^2 - yz)$. If a triple $(x, y, z)$ is a solution, then so are also the triples $(y, z, x)$ and $(z, x, y)$; thus we may assume that $x = \max\{x, y, z\}$. Then $z - x \le 0$ and $x^2 - yz \ge 0$ (remember that $x, y, z > 0$), so the equality $z - x = y(x^2 - yz)$, together with the condition $y > 0$, implies that $z - x = x^2 - yz = 0$, which means that $x = y = z$. The system then reduces to the single equation $1/x^2 = 1 + 1$, which has a (unique) positive root $x = \sqrt{2}/2$.

Conclusion. The system has exactly two solutions, $x = y = z = \pm \sqrt{2}/2$.