siddharth, you asked:
Can you prove that the only possible solutions are f(x)=x and f(x)=0

I'm simply saying that f(x) = x satisfies the given criteria [i.e. xy/2 = xy/2 for all real x, y and f'(1) = 1 = 1 = f(1)] but f is neither of those two solutions you proposed above. The stuff I wrote in brackets in my post #72 was to acknowledge that f(x) = x is not differentiable at x=0, but the given criteria only require that f be diff'able at 1, not necessarily everywhere.