Example 17 Let $f(x, y, z)$ be a polynomial in $x, y, z$, and a 4th-degree polynomial in $x$, satisfying
$$
\left\{\begin{array}{l}
f\left(x, z^{2}, y\right)+f\left(x, y^{2}, z\right)=0, \\
f\left(z^{3}, y, x\right)+f\left(x^{3}, y, z\right)=0 .
\end{array}\right.
$$

Find such a polynomial $f(x, y, z)$.
(1995 Japan Mathematical Olympiad Preliminary Problem)
Given the conditions, let's consider setting $F(x, y, z) = f(x^3, y^2, z)$, then $F(x, z, y) = -F(x, y, z)$, and $F(z, y, x) = -F(z, x, y)$.
Accordingly, we have $F(y, x, z) = -F(z, x, y) = F(z, y, x) = -F(x, y, z)$.
Therefore, $F(x, y, z)$ is an alternating polynomial in $x, y, z$.
Moreover, it can be proven that the following equations hold: $F(x, y, z) = F(y, z, x) = F(z, x, y)$.
Since $F(x, y, z)$ is an alternating polynomial, there exists a symmetric polynomial $g(x, y, z)$ such that
$$
f(x^3, y^2, z) = F(x, y, z) = g(x, y, z)(x-y)(y-z)(z-x).
$$

Replacing $y$ with $-y$ in the above equation, we get
$$
f(x^3, y^2, z) = F(x, -y, z) = g(x, -y, z)(x+y)(-y-z)(z-x),
$$

which shows that $f(x^3, y^2, z)$ is also divisible by $x+y$. Similarly, let $\omega$ be a root of $\omega^3 = 1$ other than 1, and replace $x$ with $x\omega$ and $x\omega^2$, while $f(x^3, y^2, z)$ remains unchanged, indicating that $f(x^3, y^2, z)$ is divisible by $x\omega - y$, $x\omega^2 - y$, $x\omega + y$, and $x\omega^2 + y$. Therefore, $f(x^3, y^2, z)$ is divisible by $(x-y)(x\omega-y)(x\omega^2-y)(x+y)(x\omega+y)(x\omega^2+y) = (x^3 - y^3)(x^3 + y^3) = x^6 - y^6$.
Additionally, $f(x^3, y^2, z) = F(x, y, z) = F(y, z, x) = f(y^3, z^2, x)$.
Thus, $f(x^3, y^2, z)$ is divisible by $y^6 - z^6$. Similarly, it is also divisible by $z^6 - x^6$.
Therefore, there exists a symmetric polynomial $h(x, y, z)$ such that
$$
f(x^3, y^2, z) = h(x, y, z)(x^6 - y^6)(y^6 - z^6)(z^6 - x^6).
$$

Hence, $(x^6 - y^6)(y^6 - z^6)(z^6 - x^6)$ is a factor of $f(x^3, y^2, z)$. At this point, $f(x, y, z)$ is a polynomial of degree 4 in $x$, so one possible answer is $(x^2 - y^3)(y^3 - z^6)(z^6 - x^2)$.