>>8
Looked at my old notes from analysis last year.
Apparently I have proven that if f(x) = x + 2x^2sin(1/x) if x!=0 and f(0)=0
then f'(0) = 1, but there is no interval around 0 where f is increasing, which means it can't be continous.
So I define my function as
f(x,y) = x + y + 2x^2sin(1/x) + 2y^2sin(1/y)
f(0,y) = y + 2y^2sin(1/y)
f(x,0) = x + 2x^2sing(1/x)
f(0,0) = 0
And if my result is true then that should satisfy your conditions :p
I know it sounds a bit weak, but I could prove that first result if I could be arsed, I hate typing maths into textboxes like thes :p