I'm dubious in step one proof. The reference I'm following claims to use the Implicit Function Theorem, but I haven't seen where it was used.
Lagrange Multipliers Theorem: Let $f:U\to\mathbb{R}$ a differential function in a open $U \subset \mathbb{R}^{m+n}$ and $M = \varphi^{-1}(c)$ a orietable surface in $U$, inverse image of regular value $c \in \mathbb{R}^n$ by a aplication $\varphi:U \to \mathbb{R}^n$, of class $C^1$. A point $p \in M$ is critic of restriction $f|M$ iff there exists $\lambda_1, \ldots, \lambda_n$ such that $$\nabla f(p) = \lambda_1 \cdot \nabla \varphi_1 (p) + \ldots + \lambda_n \cdot \nabla \varphi_n(p),$$ where $\varphi_1, \ldots, \varphi_n: U \to \mathbb{R}$ are the coordinates of $\varphi$.
Proof: Assume the notation of the statement. Since $c$ is a regular value of $\varphi$, then for each $x \in M$, the vectors $\nabla \varphi_1 (x) ,\ldots, \nabla \varphi_n(x)$ are linearly independent and so this form a basis of vector space normal in $M$ in the point $x$, denoted by $(T_xM)^{\bot}$. In this case, if $f:U \to \mathbb{R}$ a differentiable function, then $p \in M$ is a critic point of the $f|M$ iff the vector $\nabla f(p)$ is a linear combination of vectors $\varphi_i (p)$, ie, there exists $\lambda_1, \ldots, \lambda_n$ such that $$\nabla f(p) = \lambda_1 \cdot \nabla \varphi_1 (p) + \ldots + \lambda_n \cdot \nabla \varphi_n(p).$$
I can't figure out where he used the implicit function theorem, I believe it's to show that those vectors form the basis, but I'm not convinced of that. I looked at the case in one variable, it seems to be in that step. Thank you in advance for the tips.
Next I did a proof trying to explain where the theorem was used.
Proof: If $x \in M$, then by $M = \varphi^{-1}(c)$, it follows that $\varphi(x) = c$ and so, since $c$ is a regular value of $\varphi=(\varphi_1, \ldots, \varphi_n):U \to \mathbb{R}^n$, then $\nabla \varphi_i(x) \neq 0$, for all $i = 1 ,\ldots, n$. Thus, by the implicit function theorem, there exists a open $Z \subset U$ with $x \in Z$ such that $\varphi^{-1}(c) \cap Z$ is the graphic of function $\xi:V \to \mathbb{R}^n$, o class $C^k$ in a open $V \subset \mathbb{R}^n$. Thus, considering the open $V$, we have that the set $\{\nabla \varphi_1 (x), \ldots, \nabla \varphi_n(x)\}$ is linearly independent and so this form a basis of vector space normal in $M$ in the point $x$, denoted by $(T_xM)^{\bot}$. In this case, if $f:U \to \mathbb{R}$ a differentiable function, then $p \in M$ is a critic point of the $f|M$ iff the vector $\nabla f(p)$ is a linear combination of vectors $\varphi_i (p)$, ie, there exists $\lambda_1, \ldots, \lambda_n$ such that $$\nabla f(p) = \lambda_1 \cdot \nabla \varphi_1 (p) + \ldots + \lambda_n \cdot \nabla \varphi_n(p).$$
Is this second version correct? I'm not convinced it's right.
In short, my question is: at what point is the Implicit Function Theorem being used.
Taking the question, I would like to ask what would be another interesting application of this theorem in the context of analysis in $\mathbb{R}^n$.