So just based on the cauchy riemann theorem, I think:
Ux = 2 = Vy = 2xy, so f(z) is differentiable on xy = 1, and also that Vx = y^2 = -Uy = 0. That doesn't make sense to me because if 0 = y^2, then y = 0, yet that wouldn't satisfy xy = 1, would it?
Furthermore, I'm not sure how I would...