Czech-Slovak-Polish Match

From the form of the equations it is immediate that $xyz\ne 0$. 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, $x=y=z=\pm \sqrt{2}/2$.