The integral is path-independent if there is a scalar function [tex]f[/tex] whose gradient is
[tex]\nabla f=(2e^x\cos2y,-\sin2y)[/tex]
(at least, that's what it looks like the given integrand is)
Then
[tex]\dfrac{\partial f}{\partial x}=2e^x\cos 2y\implies f(x,y)=2e^x\cos2y+g(y)[/tex]
Differentiating both sides with respect to [tex]y[/tex] gives
[tex]\dfrac{\partial f}{\partial y}=-4e^x\sin 2y\neq-\sin2y[/tex]
so the line integral *is* dependent on the path. (again, assuming what I've written above actually reflects what the question is asking)