Example 9 Solve the system of equations
$$
\left\{\begin{array}{l}
x^{3}=y^{2}+y+\frac{1}{3}, \\
y^{3}=z^{2}+z+\frac{1}{3} \\
z^{3}=x^{2}+x+\frac{1}{3} .
\end{array}\right.
$$
【Analysis and Solution】It is easy to know that $x, y, z > 0$.
Let the function $f(t) = t^2 + t + \frac{1}{3}$, which is monotonically increasing on the interval $(0, +\infty)$.
Then the system of equations becomes
$$
\left\{\begin{array}{l}
x^3 = f(y), \\
y^3 = f(z), \\
z^3 = f(x).
\end{array}\right.
$$

If $x > y$, then
$$
\begin{array}{l}
f(x) > f(y) \Rightarrow z^3 > x^3 \Rightarrow z > x \\
\Rightarrow f(z) > f(x) \Rightarrow y^3 > z^3 \Rightarrow y > z.
\end{array}
$$

Therefore, $x > y > z > x$, which is a contradiction.
If $x < y$, a similar contradiction is derived.
Thus, $x = y = z$.
Hence, $x^3 = x^2 + x + \frac{1}{3} \Leftrightarrow 4x^3 = (x + 1)^3$.
So, $x = y = z = \frac{1}{\sqrt[3]{4} - 1}$.