Answer
$(y+2)^2=-8(x+1)$, $(-3,-6),(-3,2)$.
See graph.
Work Step by Step
1. Given focus $(-3,-2)$ and directrix $x=1$ (opens leftward), we have $p=(1+3)/2=2$, vertex $(-1,-2)$, and equation (as in a standard form) $(y+2)^2=-4(2)(x+1)$ or $(y+2)^2=-8(x+1)$
2. Let $x=-3$, we have $(y+2)^2=-8(-3+1)$ and $y=-2\pm4$, thus we have the two points that define the latus rectum $(-3,-6),(-3,2)$
3. See graph.