I'm trying to solve a question using IFT and I ran into a bit of a confusing part.
We have a function $\psi(u, v): \mathbb R^2 \to \mathbb R$ which is $C^1$ and we know that $\psi(0, 0) = 0$, furthermore we also know that $\frac{\partial \psi}{\partial u}(0, 0) + \frac{\partial \psi}{\partial v}(0, 0) \neq 0$
We define another function $F: \mathbb R^3 \to \mathbb R$, defined by $F(x, y, z) = \psi(x+z, y+z)$.
Show that $F(x, y, z) = 0$ implicity defines $z = z(x,y)$ at a neighborhood of $(0, 0)$.
Where is the issue: We want to use the IFT, so we derive $F$ wrt $z$ and get:
$\frac{\partial F}{\partial z} = \frac{\partial \psi}{\partial u} + \frac{\partial \psi}{\partial v}$
We know that $\frac{\partial \psi}{\partial u} + \frac{\partial \psi}{\partial v} \neq 0$ when $(u, v) = (0, 0)$, but that's not necesarily the case.
"Show that $F(x, y, z) = 0$ implicity defines $z = z(x,y)$ at a neighborhood of $(0, 0)$." implies that $(x, y) = (0, 0)$. Not $(u, v)$.
And we can have a situation where $(x, y) = (0, 0)$ and $(u, v) \neq (0, 0)$. For instance:
if $(x, y, z) = (0, 0, 1)$ then $(u, v) = (1, 1)$
And now we have that $\frac{\partial F}{\partial z}(0, 0, 1) = \frac{\partial \psi}{\partial u}(1, 1) + \frac{\partial \psi}{\partial v}(1, 1)$ and maybe that's zero. We don't know. We weren't given any information about what happens when $(u, v) = (1, 1)$. And if that's zero, then there is no implicitly defined function necessarily.
I feel like I'm missing something.
I think your problems should read as follows:
"Show that $F(x,y,z)=0$ implicitely defines a function $z=z(x,y)$ at a neigborhood of $(0,0)$ with $z(0,0)=0$."