Example 11. Find the real solutions of the system of equations
$$
\left\{\begin{array}{l}
1-x^{2}=y, \\
1-y^{2}=z, \\
1-z^{2}=x
\end{array}\right.
$$
(i) If $x=y=z$, it is easy to get
$$
x=y=z=\frac{-1 \pm \sqrt{5}}{2} .
$$
(ii) If two of the unknowns are equal, let's assume $x=y$, we can similarly get $x=y=z=\frac{-1 \pm \sqrt{5}}{2}$.
(iii) If the three unknowns are not equal, let's assume $x>y>z$, then
$1-x^{2}=y>z=1-y^{2}$, which implies $x^{2}y$, hence $yz$, so $y>0$. This is a contradiction.
Therefore, when $x, y, z$ are not equal, the original system of equations has no real solutions. Hence, the real solutions of the original system of equations are
$$
\begin{array}{l}
x_{1}=y_{1}=z_{1}=\frac{-1+\sqrt{5}}{2} ; \\
x_{2}=y_{2}=z_{3}=\frac{-1-\sqrt{5}}{2} .
\end{array}
$$