Example 5 Given $x y z=1, x+y+z=2, x^{2}+$ $y^{2}+z^{2}=16$. Then $\frac{1}{x y+2 z}+\frac{1}{y z+2 x}+\frac{1}{z x+2 y}=$
Solution: Squaring both sides of $x+y+z=2$ yields
$$
x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=4 \text {. }
$$

Substituting $x^{2}+y^{2}+z^{2}=16$ gives
$$
x y+y z+z x=-6 \text {. }
$$

From $x+y+z=2$, we get $z=2-x-y$. Therefore,
$$
\begin{array}{l}
\frac{1}{x y+2 z}=\frac{1}{x y-2 x-2 y+4} \\
=\frac{1}{(x-2)(y-2)} .
\end{array}
$$

Similarly, $\frac{1}{y z+2 x}=\frac{1}{(y-2)(z-2)}$,
$$
\frac{1}{z x+2 y}=\frac{1}{(z-2)(x-2)} \text {. }
$$

Thus, $\frac{1}{x y+2 z}+\frac{1}{y z+2 x}+\frac{1}{z x+2 y}$
$$
\begin{array}{l}
=\frac{1}{(x-2)(y-2)}+\frac{1}{(y-2)(z-2)}+\frac{1}{(z-2)(x-2)} \\
=\frac{z-2+x-2+y-2}{(x-2)(y-2)(z-2)} \\
=\frac{x+y+z-6}{x y z-2(x y+y z+z x)+4(x+y+z)-8} \\
=\frac{2-6}{1-2 \times(-6)+4 \times 2-8}=-\frac{4}{13} .
\end{array}
$$