Example 2 Given
$$
\left(x+\sqrt{x^{2}+2000}\right)\left(y+\sqrt{y^{2}+2002}\right)=2002 \text {. }
$$

Then $x^{2}-3 x y-4 y^{2}-6 x-6 y+58=$ $\qquad$
(2002, Junior High School Mathematics Competition, Jiangsu Province)
Explain: Construct the rationalizing factors $x-\sqrt{x^{2}+2002}$ and $y-\sqrt{y^{2}+2002}$ for $x+\sqrt{x^{2}+2002}$ and $y+\sqrt{y^{2}+2002}$. Because
$$
\begin{array}{l}
\left(x+\sqrt{x^{2}+2002}\right)\left(x-\sqrt{x^{2}+2002}\right) . \\
\quad\left(y+\sqrt{y^{2}+2002}\right)\left(y-\sqrt{y^{2}+2002}\right) \\
=(-2002) \times(-2002)=2002^{2}, \\
\left(x+\sqrt{x^{2}+2002}\right)\left(y+\sqrt{y^{2}+2002}\right) \\
=2002,
\end{array}
$$
$$
\begin{array}{l}
\text { Hence }\left(x-\sqrt{x^{2}+2002}\right)\left(y-\sqrt{y^{2}+2002}\right) \\
=2002 \text {. }
\end{array}
$$

Thus, $x+\sqrt{x^{2}+2002}=\frac{2002}{y+\sqrt{y^{2}+2002}}$
$$
\begin{array}{l}
=-\left(y-\sqrt{y^{2}+2002}\right), \\
x-\sqrt{x^{2}+2002}=-\left(y+\sqrt{y^{2}+2002}\right) .
\end{array}
$$

Adding the two equations gives $2 x=-2 y$, i.e., $x+y=0$.
Therefore, $x^{2}-3 x y-4 y^{2}-6 x-6 y+58$
$$
=(x-4 y)(x+y)-6(x+y)+58=58 \text {. }
$$