Example 4 Factorize $f(x, y, z)=\left(y^{2}-z^{2}\right)(1+x y)(1+x z)+\left(z^{2}-x^{2}\right)(1+y z)(1+$ $x y)+\left(x^{2}-y^{2}\right)(1+y z)(1+x z)$.
Let $f(x, y, z)$ be a non-homogeneous cyclic polynomial of degree six in $x, y, z$. When $y=z$, the original expression $=0$. According to the cyclic property, the original polynomial has a factor $(y-z)(z-x)(x-y)$, which is a homogeneous cyclic polynomial of degree three. Therefore, the other factor is a non-homogeneous cyclic polynomial of degree three. Hence, we can set
$$
f\left(x_{1}, y, z\right)=(y-z)(z-x)(x-y) \cdot\left[k_{1}\left(x^{3}+y^{3}+z^{3}\right)+k_{2}\left(x^{2} y+y^{2} z+z^{2} x\right)+k_{3}\left(x y^{2}+y z^{2}+z x^{2}\right)+k_{4} x y z+k_{5}\left(x^{2}+y^{2}+z^{2}\right)+k_{6}(x y+y z+z x)+k_{7}(x+y+z)+k_{8}\right].
$$

By comparing the coefficients of the highest power of $x$ in the assumed form and the original polynomial, we get $k_{1}=k_{2}=k_{3}=k_{5}=0$. Thus, the above expression simplifies to
$$
f(x, y, z)=(y-z)(z-x)(x-y) \cdot\left[\left(k_{4} x y z+k_{6}(x y+y z+z x)\right.\right.
$$
$$
\left.+k_{7}(x+y+z)+k_{8}\right].
$$

By comparing the coefficients of the $x^{3} y^{2}$ term in the above expression and the original polynomial, we get $k_{6}=0$. By comparing the coefficients of the $x^{3} y$ term, we get $k_{7}=1$. By comparing the coefficients of the $x^{2} y$ term, we get $k_{8}=0$. By setting $z=1, y=2, x=3$, we get $k_{4}=1$. Therefore,
$$
f(x, y, z)=(y-z)(z-x)(x-y)(x y z+x+y+z).
$$