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 $\nabla f(\vec x_0)=\lambda \nabla g(\vec x_0)$.