Answer
asymptotes: $y=\pm\frac{2}{3}(x-3)+1$
foci: $(3,1\pm\sqrt {13})$
Work Step by Step
Step 1. Rewriting the equation as
$9(y^2-2y+1)-4(x^2-6x+9)=63+9-36=36$
or
$\frac{(y-1)^2}{4}-\frac{(x-3)^2}{9}=1$
we have $a=2,b=3, c=\sqrt {a^2+b^2}=\sqrt {13}$ centered at $(3,1)$ with a vertical transverse axis.
Step 2. We can find the vertices as $(3,-1),(3,3)$ and asymptotes as $y-1=\pm\frac{a}{b}(x-3)$ or $y=\pm\frac{2}{3}(x-3)+1$
Step 3. We can graph the equation as shown in the figure with foci at $(3,1\pm\sqrt {13})$
