Answer
$(x+2)^2-\frac{(y+3)^2}{3}=1$, See graph.
Work Step by Step
1. Based on the given conditions, the hyperbola has a horizontal transverse axis with center $(-2,-3)$, thus we have a general form $\frac{(x+2)^2}{a^2}-\frac{(y+3)^2}{b^2}=1$
2. With $c=2, a=1$ (distances to the center), we have $b^2=c^2-a^2=3$, thus the equation $(x+2)^2-\frac{(y+3)^2}{3}=1$ with asymptotes $y+3=\pm\frac{b}{a}(x+2)=\pm\sqrt 3(x+2)$
3. See graph.
