I have to show that the equation $\displaystyle z^3 + ze^{(x+y)} + 2 = 0$ has a unique solution $\displaystyle z = f(x,y)$ for all (x,y) in $\displaystyle R^2$. 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?