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

Given $x = t^3$ and $y = t^2$.

Since both $x$ and $y$ are in parametric form:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Finding the derivatives with respect to $t$:

$\frac{dy}{dt} = 2t$

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

Therefore:

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

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

---

For the second derivative in parametric form:

$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx}$

Rewriting $\frac{dy}{dx} = \frac{2}{3t} = \frac{2}{3} \cdot t^{-1}$

Differentiating with respect to $t$:

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

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

---

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

$\frac{dt}{dx} = \frac{1}{3t^2}$

---

Combining the results:

$\frac{d^2y}{dx^2} = -\frac{2}{3t^2} \times \frac{1}{3t^2}$

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