Denote and suppose . Then,
So, for every there exists in a neigborhood of a function (of class 1) satisfying and for all .
I have to show that the equation has a unique solution for all (x,y) in . Conclude from the Implicit Function Theorem that f is continuous everywhere.
If I solve for z=f(x,y) in a neighbourhood of (0,0,-2) is this the same thing as showing that z=f(x,y) for all (x,y)?
Or is there another way to do it without using successive approximations?