Exercise 7 in section 14.3 in Fitzpatrick's Calculus says:
Suppose that the function $f: \mathbb{R}^n \to \mathbb{R}$ has continuous second-order partial derivatives. Let $x$ be a point in $\mathbb{R}^n$ at which $\nabla f(x) = 0$. Assume also that there are points $u$ and $v$ in $\mathbb{R}^n$ at which $\left<\nabla^2 f(x) u, u \right> > 0$ and $\left<\nabla^2 f(x) v,v \right> < 0$. Show that the point $x$ is neither a local maximum nor a local minimum of the function $f: \mathbb{R}^n \to \mathbb{R}$.
I think this question is wrong:
When $\left<\nabla^2 f(x) u, u \right> >0$ for some $u$ then it is positive for some neighborhood of $u$ as second-order partial derivatives and their scalar products are continuous; and its behavior is irrelevant to another (separate) point, say $v$. Am I right?
No, I think you're confusing $x$ and $u$. We're interested in values in a neighbourhood of $x$. For $u$, neighbourhoods are irrelevant; both $u$ and $v$ are direction vectors along which you test the derivatives of $f$. The formulation of the exercise is a bit confusing; "there are points $u$ and $v$ in $\mathbb R^n$" invites this misunderstanding. I wouldn't call $u$ and $v$ "points" here, much like one wouldn't call a velocity vector a "point" even though of course technically it's correct that it's a "point" in $\mathbb R^n$. The vectors $u$ and $v$ live in a different $\mathbb R^n$ from $x$ much as the velocity vector lives in a different $\mathbb R^n$ from the position vector.