I am working on the following exercise : we consider the function $f(x, y) = 2x^3 - 5xy + 3y^2$ from $\Bbb R^2$ to $\Bbb R$ and would like to find the values $c \in \Bbb R$ such that the set $f^{-1}(c)$ is a submanifold of $\Bbb R^2$.
I have only one theorem in my lecture notes about that and it tells :
Let $f : U \subset \Bbb R^m \to \Bbb R^n$ a $C^k$ function of constant rank $=r$ then for each point $q \in \Bbb R^n$, the preimage $M = f ^{-1}(q) \subset U$ is a differentiable submanifold of dimension $m - r$.
The solution of the exercise uses the following result : if $f : U \subset \Bbb R^n \to \Bbb R$ is $C^k$ and if $df_p \neq 0$ for all $p$ such that $f(p) = c$, then $f^{−1}(c)$ is a $C^1$ submanifold. To me, that is not the same theorem we stated. In fact the function $f$ is not of constant rank on the whole $\Bbb R^2$, there are a couple of points where the gradient is $0$ so we cannot use the theorem. So the result they are using is something different, it is a corollary from the theorem ? Did I misunderstand the theorem ?
Define a family of subsets of $\mathbb R^2$, given by $$M_c:=\{(x,y)\in\mathbb R^2 | f(x,y)=c\}.$$ We want to find the values of $c$ such that $M_c$ defines a submanifold of $\mathbb R^2$.
First, we have to find the regular values of the smooth map $f:\mathbb R^2\to\mathbb R$, where $f(x,y)=2x^3-5xy+3y^2$ (observe that $x,y$ are the global coordinates of the real plane).
A regular value is a point $q\in f(\mathbb R^2)$ such that $\forall p\in f^{-1}(\{q\})$, the map $f_{*p}:T_p\mathbb R^2\to T_{f(p)}\mathbb R$ is surjective.
The differential map $f_{*p}$ is linear and its matrix representation is given by $$\nabla f(x,y)\vert_p=(6x^2-5y,-5x+6y)\vert _p$$ whose rank is less than one iff $\begin{cases}6x^2-5y=0\\-5x+6y=0 \end{cases}\iff \begin{cases}6x^2-5y=0\\ x=(6/5)y\end{cases}\iff 6(6y/5)^2-5y=0$.
Now you can find the solutions for the system to see for which $c\in\mathbb R$, the set $M_c$ is a $2-1$-dimensional submanifold of $\mathbb R^2$.