Vector perpendicular to level set of objective function

Question

A constant level contour of an objective function $ \mathbb{R^n} \rightarrow \mathbb{R}$, corresponding to a given value c is the set: $$ F_c(c)= \{ x \in \mathbb{R}^n | f(x)=c \} $$ Given points $x_0, x_1 \in \mathbb{R}^n $ suppose that the vector $v = x_1-x_0$ is perpendicular to $F_c$ at the point $x_0$. What does this mean in strict mathematical language ? Does the function $F$ need to be differentiable at $x_0$? How can we prove that gradient whenever exists at $x_0$ is vector perpendicular to $F_c$ at $x_0$ and to points towards increasing values of $f$?

Answer 1

Well, $\frac{df(x)}{x}$ is the gradient.
The vectors perpendicular to the gradient are the direction that $f(x) = const.$
Don't know if this helps.