If $x=at^2$, y = 2at,  then $\frac{d^2 y}{d x^2}$ is :

$-\frac{1}{2at^3}$

$x=at^2 \Rightarrow \frac{d x}{d t}=2 a t$

$y=2at \Rightarrow \frac{d y}{d t}=2 a$

∴  $\frac{d y}{d x}=\frac{2 a}{2 a t}=\frac{1}{t} \Rightarrow \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{1}{t}\right)=-\frac{1}{t^2} \frac{d t}{d x}$

$=\left(-\frac{1}{t^2}\right)\left(\frac{1}{2 a t}\right)$     [By (1)]

$=-\frac{1}{2 a t^3}$

Hence (4) is correct answer.