Second derivative of a parametric curve

First compute the derivatives with respect to $t$: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2$. For $t \neq 0$, we have $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3t}{2}$. Note that at $t = 0$, we have $\frac{dx}{dt} = 0$, so the derivative is undefined there; the curve has a cusp at this point.

Now differentiate $\frac{dy}{dx} = \frac{3t}{2}$ with respect to $t$, giving $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2}$. For $t \neq 0$, dividing by $\frac{dx}{dt} = 2t$ gives $\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}$. This formula is valid for all $t \neq 0$, and is undefined at $t = 0$ due to the cusp in the parametric curve.