Answer
The upper half of the circle centered at $(0,-3)$, with radius $3.$
The circle and the diameter are excluded.
Work Step by Step
$ x^{2}+y^{2}+6y=0\quad$ is a border curve for the first region
Complete the square:
$x^{2}+(y^{2}+2\cdot y\cdot 3+3^{2}-3^{2})=0$
$x^{2}+(y+3)^{2}-9=0$
$x^{2}+(y+3)^{2}=9$
which is a circle centered at $(0,-3)$, with radius $3.$
The second equation restricts the y-coordinate
to be above $-3$.
So the region is:
The upper half of the circle centered at $(0,-3)$, with radius $3.$
The circle and the diameter are excluded.
