Example 1 Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c=-1, a+b+c=4, \\
\frac{a}{a^{2}-3 a-1}+\frac{b}{b^{2}-3 b-1}+\frac{c}{c^{2}-3 c-1}=1 .
\end{array}
$$

Find the value of $a^{2}+b^{2}+c^{2}$.
From the conditions given in the problem, we have
$$
\frac{1}{a}=-b c, \quad a=4-b-c \text {. }
$$

Notice that,
$$
\begin{array}{l}
\frac{a}{a^{2}-3 a-1}=\frac{1}{a-3-\frac{1}{a}}=\frac{1}{b c-b-c+1} \\
=\frac{1}{(b-1)(c-1)} .
\end{array}
$$

Similarly, $\frac{b}{b^{2}-3 b-1}=\frac{1}{(c-1)(a-1)}$,
$\frac{c}{c^{2}-3 c-1}=\frac{1}{(a-1)(b-1)}$.
Therefore, $\frac{a}{a^{2}-3 a-1}+\frac{b}{b^{2}-3 b-1}+\frac{c}{c^{2}-3 c-1}$
$=\frac{1}{(b-1)(c-1)}+\frac{1}{(c-1)(a-1)}+\frac{1}{(b-1)(c-1)}$
$=\frac{a+b+c-3}{(a-1)(b-1)(c-1)}$
$=\frac{1}{(a-1)(b-1)(c-1)}=1$
$\Rightarrow (a-1)(b-1)(c-1)$
$=a b c-(a b+b c+c a)+(a+b+c)-1$
$=1$
$\Rightarrow a b+b c+c a=1$
$\Rightarrow a^{2}+b^{2}+c^{2}$
$=(a+b+c)^{2}-2(a b+b c+c a)=14$.