Find the equation

finding the slope of the tangent by differentiating the curve

$2 a y \frac{d y}{d x}=3 x^{2}$

$\frac{d y}{d x}=\frac{3 x^{2}}{2 a y}$

$\mathrm{m}$ (tangent) at $\left(\mathrm{am}^{2}, \mathrm{am}^{3}\right)$ is $\frac{3 \mathrm{~m}}{2}$

normal is perpendicular to tangent so, $m_{1} m_{2}=-1$

$\mathrm{m}$ (normal) at $\left(\mathrm{am}^{2}, \mathrm{am}^{3}\right)$ is $-\frac{2}{3 \mathrm{~m}}$

equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$

$y-a m^{3}=-\frac{2}{3 m}\left(x-a m^{2}\right)$