If x=t^2 and y=t^3-3t for t>0, then (d^2y)/(dx^2 at t=2 equals

First find $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2-3}{2t}=\dfrac32\left(t-\dfrac1t\right)$. Differentiate this with respect to $x$ by the chain rule: $\dfrac{d^2y}{dx^2}=\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)\Big/\dfrac{dx}{dt}=\dfrac32\left(1+\dfrac1{t^2}\right)\cdot\dfrac1{2t}=\dfrac{3(t^2+1)}{4t^3}$. At $t=2$ this is $\dfrac{3\cdot5}{4\cdot8}=\dfrac{15}{32}$. The usual error is to omit the final division by $dx/dt$, i.e. to differentiate with respect to $t$ only. A quick numerical check: at $t=2$ we have $\dfrac{dy}{dx}=\dfrac32\left(2-\dfrac12\right)=\dfrac94$, and the slope increases with $t$, consistent with a positive second derivative. Using the formula $\dfrac{3(t^2+1)}{4t^3}$ at $t=1$ gives $\dfrac32$, matching direct differentiation of the parametric forms.