If x=t^3, y=t^4 then (d^2 y)/(d x^2) at t=2 is

$x=t^3$           $y=t^4$

$\frac{d x}{d t}=3 t^2$       $\frac{d y}{d t}=4 t^3$

$\Rightarrow \frac{d y / d t}{d x / d t}=\frac{4 t^3}{3 t^2}$

$\Rightarrow \frac{d y}{d x}=\frac{4 t}{3}$

so $\frac{d^2 y}{d x^2}=\frac{4}{3} \frac{d t}{d x}$

$\Rightarrow \frac{d^2 y}{d x^2}=\frac{4}{3 \times 3 t^2}=\frac{4}{9 t^2}$

so at t = 2

$\Rightarrow \frac{4}{9 \times(2)^2}$

$=\frac{1}{9}$