Answer
This is an ellipse, with the major axis on the $y$-axis:
the center is $(0,-3)$
the foci are $(0, -3\pm\sqrt{3})$
the vertices are $(0,0)$ and $(0,-6)$
Work Step by Step
We group like terms
$9x^{2}+(6y^{2}+36y)=0$
$9x^{2}+6(y^{2}+6y)=0 \quad $
We complete the square
$9x^{2}+6(y^{2}+6y+3^{2})=6(3^{2})$
$9x^{2}+6(y+3)^{2}=54\quad $
Divide with $54$:
$\displaystyle \frac{x^{2}}{6}+\frac{(y+3)^{2}}{9}=1$
This has the form $\quad \displaystyle \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1 \quad (a\gt b)$
Foci on the y-axis$,\quad a=3,\quad b=\sqrt{6}$
Center-to-focus distance: $\quad c=\sqrt{a^{2}-b^{2}}=\sqrt{9-6}=\sqrt{3}$
Foci: $(0, \pm\sqrt{3})$
Vertices: $(0, \pm 3)$
When $y$ is replaced with $(y+3)$, a shift downward is made.
$(x,y)\rightarrow(x,y-3)$
(No horizontal shift $\Rightarrow$ the main axis remains on the y-axis.)
The center shifts to $(0,0-3)=(0,-3)$
The foci shift to $(0, -3\pm\sqrt{3})$
The vertices shift to $(0, -3\pm 3)\Rightarrow (0,0)$ and $(0,-6)$
