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

$\frac{-2}{9t^4}$

The correct answer is Option (4) → $\frac{-2}{9t^4}$

Given parametric equations: $x=t^3$, $y=t^2$

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

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t}{3t^2}=\frac{2}{3t}$

Now, $\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{1}{\frac{dx}{dt}}$

$\frac{d}{dt}\left(\frac{2}{3t}\right)=\frac{d}{dt}\left(\frac{2}{3}t^{-1}\right)=-\frac{2}{3}t^{-2}$

So, $\frac{d^2y}{dx^2}=\frac{-\frac{2}{3t^2}}{3t^2}=-\frac{2}{9t^4}$

Final Answer: $\frac{d^2y}{dx^2}=-\frac{2}{9t^4}$