Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ be defined by:
$(x,y)\in \mathbb{R}^2 \rightarrow f(x,y)= x^2 - y^2$.
Find critical points, regular value, regular points.
My attempt:
$[D(a,b)f]=[2a -2b]$. This has rank $1$ since the number of non-zero rows is $1$, except at $(a,b)={\{\vec0\}}$. $a\in \mathbb{R}^2$ is a critical point of $D(a,b)f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ if it has rank less than $m$. Here, $n=2,m=1$. ${\{\vec0\}}$ has rank $0<m$, so ${\{\vec0\}}$ is a critical point of $f$.
A regular value of $f$ is a point in $\mathbb{R}^m-C_f$, here $C_f$ denotes the critical points of $f$. So in this case, the regular value of $f$ (denoted by $R_f$) is $R_f=\mathbb{R}-{\{\vec0\}}$.
The set of regular points of $f$ is $f^{-1}(R_f)$ so .. it is $f^{-1}(0)$, which is a manifold since,by definition, the inverse image of a regular value is a manifold.
If my attempt is correct, is there a way to calculate $f^{-1}(0)$?