If you have dx/dt = some function in terms of displacement, x, etc., then I believe you would flip both sides of the equation, i.e. get dt/dx = 1/some function of x, then integrate with respect to x. So in this question,
dx/dt = e-2x ms-1
dt/dx = 1/e-2x
= e2x
Therefore t = Int. [e2x] dx...