China Mathematical Competition

If $z = 0$, then
$$\begin{cases} x + y = 0, \\ xy + y = 0 \end{cases}.$$
It follows that
$$\begin{cases} x = 0, \\ y = 0 \end{cases} \text{ or } \begin{cases} x = -1, \\ y = 1. \end{cases}$$
If $z \neq 0$, from $xyz + z = 0$ we get
$$xy = -1. \qquad \textcircled{1}$$
From $x + y + z = 0$ we have
$$z = -x - y. \qquad \textcircled{2}$$
Substituting $\textcircled{2}$ into $xy + yz + xz + y = 0$, we obtain
$$x^2 + y^2 + xy - y = 0. \qquad \textcircled{3}$$
From $\textcircled{1}$ we have $x = -\frac{1}{y}$. We substitute it into $\textcircled{3}$, and then make a simplification:
$$(y - 1)(y^3 - y - 1) = 0.$$
It is easy to see that $y^3 - y - 1$ has no rational solution, and so $y = 1$. Then from $\textcircled{1}$ and $\textcircled{2}$ we get $x = -1$ and $z = 0$, contradicting $z \neq 0$.

In summary, the system has exactly two solutions:
$$\begin{cases} x = 0, \\ y = 0, \\ z = 0 \end{cases}, \begin{cases} x = -1, \\ y = 1, \\ z = 0 \end{cases}$$