If $x=3t, y=5y^3+6t+1, $ then the value of $\frac{d^2y}{dx^2}$ at $t=3$ is :

10

$x=3t \Rightarrow \frac{dx}{dt}=3$

$y=5t^3+6t+1 \Rightarrow \frac{dy}{dt}=15t^2+6$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{15t^2+6}{3}=5t^2+2$

$\frac{d}{dt}\left(\frac{dy}{dx}\right)=10t$

$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{10t}{3}$

At $t=3$:

$\frac{d^2y}{dx^2}=\frac{10\times3}{3}=10$

final answer: The value of $\frac{d^2y}{dx^2}$ at $t=3$ is 10