If $(x, y)$ is a solution to the system
$xy=6$ and $x^2y+xy^2+x+y=63$,
find $x^2+y^2$.
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ \frac{1173}{32} \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 69 \qquad \textbf{(E)}\ 81$
First note that $x^2y+xy^2+x+y= (xy+1)(x+y)$. Substituting $6$ for $xy$ gives $7(x+y)= 63$, giving a result of $x+y=9$. Squaring this equation and subtracting by $12$, gives us $x^2+y^2= \boxed{69}$