I would like to prove that
$$ A=\{(x,y,z) : z= \frac{\sin(x^2+y^2)}{x^2+y^2}, (x,y)\neq 0\} $$
is a smooth sub-manifold of dimension $2$.
I think it is a direct application of the inverse function theorem on the map $f(x,y,z)=(x,y,z-\frac{\sin(x^2+y^2)}{x^2+y^2})$ where $f : (\mathbb{R}^{2}\setminus\{(0,0)\})\times\mathbb{R}\to\mathbb{R}^{3}$.
Attempt :
A simple computation proves that its differential is surjective ($\det(df_{(x,y,z)})\neq 0$) at any point of $A\subset(\mathbb{R}^{2}\setminus\{(0,0)\})\times\mathbb{R}$. Let $(x,y,z)\in A$, by the inverse function theorem, there exists an open set $U$ of $(\mathbb{R}^{2}\setminus\{(0,0)\})\times\mathbb{R}$ such that $f : U\to f(U)$ is a diffeomorphism. Now let $V=U\cap A$, we have that $\tilde{f}=f_{|V}$ is an homeomorphism from $V$ to $\tilde{f}(V)$ where $\tilde{f}(V)$ is now an open set of $\mathbb{R}^{2}\times\{0\}$ (considering the subspace topology on $A$ and $\mathbb{R}^{2}\times\{0\}$) and it admits a smooth extension (i.e. a map $F$ defined on an open set $O\subset(\mathbb{R}^{2}\setminus\{(0,0)\})\times\mathbb{R}$ smooth in the classical sense such that $F_{|V}=\tilde{f}$) trivially given by $f$. The same applies for its inverse $\tilde{f}^{-1}$. Finally, $\mathbb{R}^{2}\times\{0\}$ is diffeomorphic to $\mathbb{R}^2$, denote $p : \mathbb{R}^{2}\times\{0\}\to \mathbb{R}^2$ this diffeomorphism. Then $p\circ\tilde{f}$ sends diffeomorphically $V$ an open set of $A$ containing $(x,y,z)$ to an open set of $\mathbb{R}^2$.
Since $(x,y,z)\in A$ was arbitrary, this shows that $A$ is a smooth sub-manifold of dimension $2$.
Is this seems correct to you ?
Thank you a lot !
Sorry I forgot to mention that the definitions of a smooth manifold, a smooth map and a diffeomorphism that I use are the one used in Milnor’s book.In particular Paul Frost’s answer in this post detailed it Milnor's definition of smooth manifold
