Example 22 (Mathematical Problem 1433 from "Mathematical Bulletin") Solve the system of equations in the real numbers:
$$
\left\{\begin{array}{l}
y=x^{3}(3-2 x), \\
z=y^{3}(3-2 y), \\
x=z^{3}(3-2 z) .
\end{array}\right.
$$
(1) When $x=0$, from (1) we get $y=0$, from (2) we get $z=0$.
(2) When $x \geqslant \frac{3}{2}$, from (1) we get $y \leqslant 0$, from (2) we get $z \leqslant 0$, from (3) we get $x \leqslant 0$, contradiction.
(3) When $0 < x < \frac{3}{2}$, from (1) we get $y > 0$, from (2) we get $z > 0$, from (3) we get $x > 0$, so $0 < x < \frac{3}{2}, y > 0, z > 0$.
(4) When $x > 0, y > 0, z > 0$, let $x = u^3(3 + 2u)$, $y = v^3(3 + 2v)$, $z = w^3(3 + 2w)$, where $u > 0, v > 0, w > 0$, and
$$
\left\{\begin{array}{l}
v=u^{3}(3+2 u), \\
w=v^{3}(3+2 v), \\
u=w^{3}(3+2 w) .
\end{array}\right.
$$

Let $f(u)=u^{3}(3+2 u)$, it is an increasing function when $u>0$, and $v=f(u), w=f(v), u=f(w)$. If $u, v, w$ are not all equal, assume without loss of generality that $u > v$, so $f(u) > f(v)$, i.e., $v > w$, so $f(v) > f(w)$, i.e., $w > u$, thus $u > v > w > u$, contradiction, hence $u = v = w$.

Therefore, the problem reduces to solving the equation $u = f(u) = u^{3}(3 + 2u)$, i.e., $1 = u^{2}(3 + 2u)$, or $2u^3 + 3u^2 - 1 = 0$, or $(u + 1)^2(2u - 1) = 0$. Solving this, we get $u = \frac{1}{2}$ (discard $u = -1$).
Thus, $v = w = \frac{1}{2}$, i.e., $x = y = z = -\frac{1}{2}$.
In summary, the solutions $(x, y, z)$ to the original system of equations are $(0,0,0), (1,1,1), \left(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\right)$.