The equation of the normal at the point $\left(x_{1}, y_{1}\right)$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is
$\frac{x-x_{1}}{x_{1} / a^{2}}=\frac{y-y_{1}}{y_{1} / b^{2}}$
Equation of ellipse is
$9 x^{2}+5 y^{2} =45 $
$\frac{x^{2}}{5}+\frac{y^{2}}{9} =1$
Here, $a^{2}=5,\, b^{2}=9$
Equation of normal to the ellipse at the point $(0,3)$ is
$\frac{x-0}{0 / 5}=\frac{y-3}{3 / 9} $
$\Rightarrow \, x=0$
Which is the equation of $y$ -axis. Alternative Given equation is$
9 x^{2}+5 y^{2}=45$
On differentiating, we get
$18 x+10 y \frac{d y}{d x}=0 $
$\Rightarrow \, \frac{d y}{d x}=-\frac{18 x}{10 y} $
At $(0,3),\left(\frac{d y}{d x}\right)=\frac{-18(0)}{10(3)}=0$
$\therefore $ Equation of normal is
$y-3=-\frac{1}{0}(x-0)$
$\Rightarrow \, x=0$
$\Rightarrow $ y-axis