Given equation of ellipse is
$9 x^{2}+5 y^{2}-30 y=0$
The above equation can be rewritten as
$9 x^{2}+5\left(y^{2}-6 y\right)=0$
$\Rightarrow \,9 x^{2}+5(y-6 y+9)=45$
$\Rightarrow \, 9 x^{2}+5(y-3)^{2}=45$
$\Rightarrow \, \frac{x^{2}}{5}+\frac{(y-3)^{2}}{9}=1$
Since, $a^{2}<\, b^{2}$, so axis of ellipse is $Y$ -axis.
To find vertex at $Y$ -axis, put $x=0$.
$\therefore \, 5(y-3)^{2}=45$
$\Rightarrow \, (y-3)^{2}=9$
$\Rightarrow \, y-3=\pm\, 3$
$\Rightarrow \, y=0$ or $y=6$
Therefore, tangents at vertex are $y=0, y=6$