The condition for the implicit function theorem is that the (smooth) map $f: \mathbb R^n \to \mathbb R^m$ is locally a (smooth) map of $n-k$ variables if there are locally smooth maps $g_i , i \in \{n-k+1, \dots, n\}$ with the property that $\nabla g_i$ are linearly independent.
To make this question as simple as possible let's consider maps $f: \mathbb R^2 \to \mathbb R$ so that the implicit map will be $F(x,y,z) = 0$ for some $F$. A level set of a (smooth) surface.
The condition that a gradient of a surface at a point is zero means the point in consideration is an extremum. So the condition of the theorem excludes points that are "completely flat".
But I am not sure what this means: thinking of normal extrema, like the north pole on the sphere, of course not all gradients are zero (only two(?)).
But what does a point on a surface look like if all the gradients vanish?
and also:
Why does this "flatness" prevent us from expressing one of the coordinates in terms of the other two?
Note: I assume the answer to question two will turn out to be interesting as it involves a modified version of the theorem: the theorem gives a smooth function (for the derivative of this function we need the non-zero gradients). But what happens if we drop the condition of (smoothness or) continuous differentiability and just express the last coordinate in terms of the other two so that the resulting function is maybe continuous but not differentiable?
Could someone please show me an example in which we can parameterise the surface locally as $z = f(x,y)$ where $f$ is continuous but not differentiable? (I am hoping to gain some insight from this example into why we require the gradients to be nonzero)
