Example 3 Given $x y z=1, x+y+z=2, x^{2}+$ $y^{2}+z^{2}=3$. Then
$$
\frac{1}{x y+z-1}+\frac{1}{y z+x-1}+\frac{1}{z x+y-1}
$$

is ( ).
(A) 1
(B) $-\frac{1}{2}$
(C) 2
(D) $-\frac{2}{3}$
(1997, Taiyuan Junior High School Mathematics Competition)
Given: $\because x+y+z=2$,
$$
\begin{aligned}
\therefore & x y+z-1 \\
& =x y+(2-x-y)-1 \\
& =(x-1)(y-1) .
\end{aligned}
$$

Similarly, we have
$$
\begin{array}{l}
y z+x-1=(y-1)(z-1), \\
z x+y-1=(z-1)(x-1) .
\end{array}
$$

Since $x y z=1, x^{2}+y^{2}+z^{2}=3$, then
$$
\begin{array}{l}
x y+y z+z x \\
=\frac{1}{2}(x+y+z)^{2}-\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right) \\
=\frac{1}{2} . \\
\therefore-\frac{1}{x y+z-1}+\frac{1}{y z+x-1}+\frac{1}{z x+y-1} \\
\quad=\frac{z-1+x-1+y-1}{(x-1)(y-1)(z-1)} \\
\quad=\frac{x+y+z-3}{x y z-(x y+y z+z x)+x+y+z-1} \\
\quad=-\frac{2}{3} .
\end{array}
$$