Two maps $f,g:\mathbb{R}^m\rightarrow \mathbb{R}^n$ are equivalent on $p\in \mathbb{R}^m$ if there exist smooth charts $\phi: \mathbb{R}^m\rightarrow\mathbb{R}^m,\ \varphi:\mathbb{R}^n\rightarrow\mathbb{R}^n$ s.t
$$ g = \varphi\circ f\circ \phi^{-1}$$
on a neighborhood of $p$.
The question is as follows: Consider the maps $f(x,y)=x^2+y^2,\ g(x,y)=x^2-y^2$ show that for any point $(x,y)\neq(0,0),\ f,g$ are equivalent on $(x,y)$.
Show that $f,g$ are not equivalent on $(0,0)$.
Show that even when requiring that $\varphi,\phi$ are only homeomorphisms still $f,g$ are not equivalent on $(0,0)$.
I tried using the constant rank theorem in the following way: For $(x,y)\neq(0,0)$
$$D_{(x,y)}f = (2x, 2y)$$
which has constant rank 1 on a neighborhood of $(x,y)$ then by a corollary of the constant rank theorem $\exists\phi,\varphi$ s.t
$$ \varphi\circ f\circ \phi^{-1}(z_1,z_2)=az_1+bz_2 $$ is linear. Then defining $\Phi(z_1,z_2)=\left(\frac{z_1^2}{a},-\frac{z_2^2}{b}\right)$ which gives
$$ \varphi\circ f\circ \phi^{-1}\circ\Phi (z_1,z_2) = g(z_1,z_2)$$
My problem is this doesn't seem right and I have no idea how to approach the second and third part.