Let $f(x)$ a function $f:\mathbb{R}^D \mapsto \mathbb{R}$. $g(x)$ is another $D$ dimensional function. Then we have a constraint equation $g(x)=0$. Now, we have a local, constrained extreme point $x'$ on the constraint surface $g(x)=0$, for which it is, in a close neighborhood, either $f(x') > f(y)$ or $f(x') < f(y)$ where all $y$'s are on the constraint surface as well. ($x'$ is a point which would be the aim of a Lagrange multiplier problem here).
Now, I want to show the following: At $x'$, the gradient $\nabla f(x')$ is perpendicular to all vectors which lie on the tangent plane to the surface $g(x)=0$ at point $x'$; $\nabla f(x')$ and $\nabla g(x')$ are parallel to each other. In other words, for all $\vec{u}$ with $\nabla g(x') . \vec{u}=0$ it is $\nabla f(x'). \vec{u}=0$ when $x'$ is a constrained extreme point on the surface of $g(x)=0$.
Intuitively I know that this is true, which leads to the construction of Lagrange functions. But how can this be shown true, with a rigorous proof?