Given equation of curve is
${\mathit{a}}^2={\mathit{x}}^3$
Differentiating w.r.t. $\mathrm{x}$, we get

$\Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{3{\mathit{x}}^2}{2\mathit{a}\mathit{y}}$
Slope of the tangent to the curve at $\left({\mathrm{am}}^2,{\mathrm{am}}^3\right)$ is
${\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}_{\left(\mathit{a}{\mathit{m}}^2,\mathit{a}{\mathit{m}}^3\right)}=\frac{3{\left(\mathit{a}{\mathit{m}}^2\right)}^2}{2\mathit{a}\left(\mathit{a}{\mathit{m}}^3\right)}=\frac{3{\mathit{a}}^2{\mathit{m}}^4}{2{\mathit{a}}^2{\mathit{m}}^3}=\frac{3\mathit{m}}{2}$
Slope of normal at $\left({\mathrm{am}}^2,{\mathrm{am}}^3\right)$

Equation of the normal at $\left({\mathrm{am}}^2,{\mathrm{am}}^3\right)$ is
$\mathit{y}-\mathit{a}{\mathit{m}}^3=\frac{-2}{3\mathit{m}}\left(\mathit{x}-\mathit{a}{\mathit{m}}^2\right)$
$\Rightarrow 3\mathit{m}\mathit{y}-3\mathit{a}{\mathit{m}}^4=-2\mathit{x}+2\mathit{a}{\mathit{m}}^2$
$\Rightarrow 2\mathit{x}+3\mathit{m}\mathit{y}-\mathit{a}{\mathit{m}}^2\left(2+3{\mathit{m}}^2\right)=0$