Just a review of Lagrange's Multiplier (to be corrected)

We try to create a concrete computation to find the local extreme of $  f:\mathbb{R}^n\supset U\to\mathbb{R}$ subject to the constraint $  g(\vec{x})=c$. Here $  c$ is a regular point, namely, a point such that $  g(\vec{x})=c\implies \nabla g(\vec x)\neq \vec{0}$.

Moreover, we call the surface specified by $  g(\vec x) = c$ to be $  S$, hence we are finding the local extreme of $  f$ restricted to the surface $  S$, namely, we are finding the local extreme of $  \big.f\big|_S$. Suppose now $  \big.f\big|_S$ attains the local extreme at $  \vec x = \vec x_0$, then for any $  \vec v\in\{\vec u\in\mathbb{R}^n:\nabla g(\vec x_0)\cdot \vec u = \vec 0\}:=T_{\vec x_0}S$, i.e. any tangent vector moving on $  S$ at $  \vec x_0$, we have $  D_{\vec v}f(\vec x_0)=\nabla f(\vec x_0)\cdot \vec v=0$ (for otherwise it is not an extrema).

Let's summarize the implication, $  \forall \vec v\in T_{\vec x_0}S\implies \nabla f(\vec x_0)\cdot \vec v=0$.

Since $  \nabla g(\vec x_0)$ spans the normal space of $  S$ at $  \vec x_0$, it follows that $  \nabla f(\vec x_0)=\lambda \nabla g(\vec x_0)$.