If $x = {t^2}$ and $y = {t^3},$ then $\frac{{{d^2}y}}{{d{x^2}}}$ is equal to

We have, $x = {t^2}$ and $y = {t^3}$

therefore,$\frac{{dx}}{{dt}} = 2t$ and $\frac{{dy}}{{dt}} = 3{t^2}$
therefore,$\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{3{t^2}}}{{2t}} = \frac{3}{2}t$

On further differentiating w.r.t. $x$, we get
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{3}{2} \cdot \frac{d}{{dt}}t \cdot \frac{{dt}}{{dx}}$

$= \frac{3}{2} \cdot \frac{1}{{2t}}$

$= \frac{3}{{4t}}$