For a critical point to occur, both partial derivatives must be equal to zero at that point. So the first steps would be to find the partial derivatives, and then to solve them for zero. Try that first.
\\f(x,y) = e^x(1-\cos y)
\\
\\\frac{\partial f}{\partial x} = e^x(1-\cos y),\ \ \frac{\partial f}{\partial y} = e^x\sin y
\\
\\ e^x(1-\cos y) = 0
\\ \cos y = 1
\\ y = 2k\pi,\ k\in\mathbb{Z}
\\
\\ e^x\sin y = 0
\\ \sin y = 0
\\ y = k\pi,\ k\in\mathbb{Z}
\\
\\\mbox{critical points along the lines }\ y = 2k\pi,\ k\in\mathbb{Z}
\\\mbox{which coincidentaly, is also the lines at which f(x,y) = 0}
\\
is this right
don't worry i can't work out how to use this thing bit complex anyways im pretty sure it's right i just don't know how to classify the critical points. All the critical points of the function have positive second partial derivitive in y, and zero second partial derivitive in x, and zero second partial derivitive in x and y. what kind of classification is given to that?
How about, instead of using formulas you think about what e^{x}(1- cos(y)) looks like in the vicinity of a point y= 2k[itex]\pi[/itex]? In particlar, what does 1- cos(y) look like there?
I also want to point out that
1) this is clearly homework
2) you showed NO attempt to do this in your original post.