Let $M$ be a $n$-dim manifold. If $f : M \to \mathbb{R}$ then does $\ker(df_p) = T_p(f^{-1}(c))$ for some $c \in f[M]$ This is what was written in my differential geometry class notes.

Let $M$ be a $n$-dimensional manifold. If $f : M \to \mathbb{R}$ then the subspace of $T_pM$ consisting of all the tangent vectors $X_p \in T_pM$ such that $\langle df, X \rangle = 0$ consists of all vectors tangent to the curves lying on the surface $f = \operatorname{const}$.

The way I interpreted the above was in the following way:

Let $M$ be a $n$-dimensional manifold. If $f : M \to \mathbb{R}$ then $\ker(df_p) = T_p(f^{-1}(c))$ for some $c \in f[M] \subseteq \mathbb{R}$.

Is my interpretation correct? If so how can I prove this proposition. The proof given in class relies on the definition of a tangent vector as a velocity vector of a curve I think, is there a way to view this using the definition of a tangent vector as a derivation?
 A: I suspect that there is an assumption missing in your question: observe that if $p$ is a critical point of $f$, i.e $df_p=0$, then $p$ could be an isolated local maximum or minimum, in which case you'd have a neighborhood $U$ of $p$ such that $f(q)\neq f(p)$, $\forall q\in U$. Hence, for $S=f^{-1}(f(p))$, $T_pS=0\neq T_pM=\ker df_p$. As an example for this case, take $M$ to be the sphere in $\mathbb{R}^3$ centered at $(3,0,0)$ with radius $1$, $f(x,y,z)=z$ restricted to $M$ and $p=(3,0,1)$.
It could also happen that $p$ was a saddle point of $f$, i.e $df_p=0$ but the Hessian of $f$ at $p$ has both positive and negative eigenvalues. Here, your assertion is also false. Take $M=\mathbb{R}^2$, $f(x,y)=x^2-y^2$, $p=(0,0)$. $S$ is two intersecting lines, which is not a hypersurface, so it does not even make sense to speak of tangent vectors at $p$.
However, more generally, if $p$ is a regular point of $f:M\rightarrow N$, the $S$ is a regular submanifold of codimension $\dim N$. So, because f will take any curve in $S$ through $p$ to the constant point $f(p)\in N$, its tangent vector goes to zero through $df_p$. That proves that $T_pS⊂ \ker df_p$. On the other hand, $\dim ker(df_p)=\dim M−\dim Im(df_p)=\dim M−\dim N=\dim S=\dim T_pS$. So these subspaces coincide.  
