$3 \cdot 12$ If $(x, y)$ is a solution to the system of equations
$$
\left\{\begin{array}{l}
x y=6 \\
x^{2} y+x y^{2}+x+y=63
\end{array}\right.
$$

then $x^{2}+y^{2}$ equals
(A) 13 .
(B) $\frac{1173}{32}$.
(C) 55 .
(D) 69 .
(E) 81 .
(38th American High School Mathematics Examination, 1987)
$$
\text { [Solution] From } \begin{aligned}
& x^{2} y+x y^{2}+x+y \\
= & x y(x+y)+x+y \\
= & (x+y)(x y+1),
\end{aligned}
$$

Substituting $x y=6$ into the above equation, we get $(x+y)(6+1)=63$,
$$
\therefore x+y=9 \text {, and } x^{2}+y^{2}=(x+y)^{2}-2 x y=69 \text {. }
$$

Therefore, the answer is $(D)$.
$$