If equation of normal at a point ;$\displaystyle \left ( m^{2},-m^{3} \right )$ on the curve& ;$\displaystyle x^{3}-y^{2}=0\: \: is\: \: y=3mx-4m^{3}$ then $\displaystyle m^{2}$ equals

$\displaystyle x^{3}-y^{2}=0$

$\Rightarrow \displaystyle 3x^{2}-2y\left ( \frac{dy}{dx} \right )=0$

$\Rightarrow \displaystyle 3x^{2}=2y\left ( \frac{dy}{dx} \right )$

$\Rightarrow \displaystyle \frac{3x^{2}}{2y}=\frac{dy}{dx}$

$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( m^{2},-m^{3} \right )}=\frac{3\times m^{4}}{-2\times m^{3}}$

$\therefore \displaystyle -\left ( \frac{dx}{dy} \right )=\frac{2}{3m}$

Therefore, equation of normal at given point is,

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

$\Rightarrow \displaystyle 3my+3m^{4}=2x-2m^{2}$

$\Rightarrow \displaystyle 3my=2x-3m^{4}-2m^{2}$

$\Rightarrow \displaystyle y=\frac{2x}{3m}-m^{3}-\frac{2}{3}m ..(1)$

but given equation of normal is, $\displaystyle y=3mx-4m^{3} ..(2)$

Thus comparing (1) and (2), we get

$\displaystyle 3m=\frac{2}{3m}$

$\Rightarrow \displaystyle m^{2}=\frac{2}{9}$

Hence, option 'D' is correct.