Originally Posted by
schmeling this question is a pretty straight forward use of the chain rule in higher derivatives.
There is no need for you to know the relation between y and t because as you can see in the question this relation is expressed as a differential.
what you need for solving this is to notice that if x=e^t then t=ln(x)
then dt/dx= x^(-1) and the second derivative is d^2t/dx^2= -x^(-2)
now using the chain rule
d^2y/dt^2= (d^2y/dt^2) * (dt/dx)^2 + (dy/dt) * (d^2t/dx^2)
its enough that you substitute for the values of dt/dx and d^2t/dx^2 what you derived in the beginning and you're done.
All it takes is understanding that the chain rule can be used for higher derivatives as well.