Find the equation of the normal at the point $\left( {a{m^2},\,a{m^3}} \right)$ for the curve $a{y^2} = {x^3}$.

We have,
$a{y^2} = {x^3}$ …(i)

Differentiating (i) w. r.t. $x$, we get
$a(2y)\cfrac{{dy}}{{dx}} = 3{x^2} \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{3{x^2}}}{{2ay}}$
Therefore, Slope of tangent at $\left( {a{m^2},\,a{m^3}} \right)$

$= {\left( {\cfrac{{dy}}{{dx}}} \right)_{\left( {a{m^2},\,a{m^3}} \right)}} = \cfrac{{3{{(a{m^2})}^2}}}{{2a(a{m^3})}} = \cfrac{3}{2}\,\,m$
$\Rightarrow$ Slope of normal at the given point$= - \cfrac{1}{{\cfrac{3}{2}m}} = - \cfrac{2}{{3m}}$

Hence the required equation of normal at the given point is
$\left( {y - a{m^3}} \right) = - \cfrac{2}{{3m}}\left( {x - a{m^2}} \right)$

or $3my - 3a{m^4} = - 2x + 2a{m^2}$

or $2x + 3my - 3a{m^4} - 2a{m^2} = 0$