Answer
$\frac{(x-3)^2}{9}-\frac{(y-1)^2}{4}=1$
See graph.
Work Step by Step
1. Based on the given conditions, the hyperbola has a horizontal transverse axis with center $(3,1)$ (midpoint between vertices), thus we have a general form $\frac{(x-3)^2}{a^2}-\frac{(y-1)^2}{b^2}=1$
2. With $a=3$, asymptote $3y+2x=9$, we have $-\frac{b}{a}=-\frac{2}{3}$, thus $b=2$ and $c^2=a^2+b^2=13$, thus the equation is $\frac{(x-3)^2}{9}-\frac{(y-1)^2}{4}=1$
3. See graph.
