let's look at a very simple example: when m = n = 1. define:
F(x,y) = x2+y2 - 1.
consider now (a,b) such that F(a,b) = 0. suppose that a = √2/2, for example. well, we have two choices for b, both of which work: b = √2/2, and b = -√2/2.
if we use (a,b) = (√2/2,√2/2) we get the function (defined only in a neighborhood of √2/2 in R) f(x) = √(1 - x2).
if we use (a,b) = (√2/2,-√2/2), we get the function f(x) = -√(1 - x2).
if we try to define f in a neighborhood of a = 1, we have a problem: neither half of the circle works. why is this so? let's look at the Jacobian matrix DF:
at the point (1,0), the square (1x1) matrix 2b is not invertible, we have b = 0, and the inverse of the 1x1 matrix [2b] does not exist, since det([2b]) = 2b = 2(0) = 0.
this is the important part you're missing about the implicit function theorem, there is a condition on how F is changing with respect to the y-variables, the nxn matrix (∂Fm+i/∂yj) MUST be invertible.
the second key point is "which solution" of F(x1,....,xm,y1,....,yn) = 0 we choose (and there could be several), depends on which point:
(a1,...,am,b1,....,bn) we're at, the implicitly defined function f is only defined LOCALLY (it may be impossible to find a function f that works everywhere).
here is a more "grown-up example":
let F(x,y,z,w) = (x2+y2-z2-w2,x2+y2+z+w) this is a function from R4 to R2.
clearly F(1,1,-1,-1) = (0,0). let's check the following matrix determinant:
this tells us that (1,1,-1,-1) is a "bad point" to try to implicitly define a function f with F(x,y,f(x,y)) = (0,0).
on the other hand, we also have F(√2/2,√2/2,-1,0) = (0,0) and for this point the matrix above is:
, which has non-zero determinant, so this is a "good point".