Solution:
Since $x y z \neq 0$, we can divide the second relation by the first. Observe that
$$x^{4}+x^{2} y^{2}+y^{4}=\left(x^{2}+x y+y^{2}\right)\left(x^{2}-x y+y^{2}\right)$$
holds for any $x, y$. Thus we get
$$\left(x^{2}-x y+y^{2}\right)\left(y^{2}-y z+z^{2}\right)\left(z^{2}-z x+x^{2}\right)=x^{2} y^{2} z^{2}$$
However, for any real numbers $x, y$, we have
$$x^{2}-x y+y^{2} \geq |x y|$$
Since $x^{2} y^{2} z^{2}=|x y||y z||z x|$, we get
$$|x y||y z||z x|=\left(x^{2}-x y+y^{2}\right)\left(y^{2}-y z+z^{2}\right)\left(z^{2}-z x+x^{2}\right) \geq |x y||y z||z x|$$
This is possible only if
$$x^{2}-x y+y^{2}=|x y|, \quad y^{2}-y z+z^{2}=|y z|, \quad z^{2}-z x+x^{2}=|z x|$$
hold simultaneously. However $|x y|= \pm x y$. If $x^{2}-x y+y^{2}=-x y$, then $x^{2}+y^{2}=0$ giving $x=y=0$. Since we are looking for nonzero $x, y, z$, we conclude that $x^{2}-x y+y^{2}=x y$ which is same as $x=y$. Using the other two relations, we also get $y=z$ and $z=x$. The first equation now gives $27 x^{6}=x^{3}$. This gives $x^{3}=1 / 27$ (since $x \neq 0$ ), or $x=1 / 3$. We thus have $x=y=z=1 / 3$. These also satisfy the second relation, as may be verified.