Answer
$x^2=-12y$, $(-6,-3),(6,-3)$
See graph.
Work Step by Step
1. Given the focus at $(0,-3)$, vertex at $(0,0)$, we can write the parabola equation in a standard form $4py=-x^2$. With $p=3$, we have $x^2=-12y$
2. Let $y=-3$, we have $x^2=36$ and $x=\pm6$ and the two points that define the latus rectum are $(-6,-3),(6,-3)$
3. See graph.