are non-degenerate critical points always isolated? I have a question regarding the isolation of critical points of a function:

Suppose $f : \mathbb{R}^n \to \mathbb{R}$ is a $C^\infty$ function such that $f$ has a non - degenerate critical point at $0 \in \mathbb{R}^n$. That is, we have $\triangledown f (0) = 0$, and the Hessian $\left((\partial^2 f/ \partial x_i \partial x_j ) (0) \right)$ is invertible.

Can I deduce from this that the critical point at $0$ is an isolated critical point ? My guess is to say yes, because the fact that the Hessian is non-degenerate forces $f$ to change its value in the vicinity, and in all directions. But I am unsure, in particular with regards to the last statements ("in all directions") - though that should be encoded by the fact all eigenvalues of the Hessian are non - zero.

Is this the right way to think about the question of whether the critical point is isolated ?
Thanks for your thoughts !
Yes, this is a right way of thinking about the question.  You can see this in several ways.  

1: The Morse lemma:  The Morse lemma declares that about any nondegenerate critical point $p$ of a smooth function $f$ there is a coordinate neighborhood $x_i$ so that in those coordinates, $f(x_i) = x_1^2 + \cdots + x_i^2 - x_{i+1}^2 - \cdots - x_n^2 + f(p)$.  It's an easy exercise to compute the gradient in these coordinates, and you can immediately see that the only zero of the gradient in the neighborhood is at $p$.  (In this case, $p = 0$ and the claim is that you can precompose $f$ with a diffeomorphism $\varphi$ of $\mathbb{R}^n$ so that $f\circ\varphi = \sum x_i^2 - \sum x_j^2$.)

2: Directly:  Take a curve (easiest, a straight line) $\gamma$ through $0$, and for simplicity's sake, parametrize the curve so that $\gamma(0)=0$.  Consider $f|_\gamma$.  Then $f'(t) = (\nabla f\cdot \gamma')(t)$ and $f''(t) = (\gamma'^THess_f\gamma')(t)$.  Leaving out some details you should verify as an exercise, the Hessian is invertible, so in a neighborhood of $0$, $f$ has a local extremum, hence the zero of $f'$ is isolated.  This is true for any curve, thus, every direction, so $0$ is an isolated critical point.  This formalizes your intuition about "every direction."
In Neal answer ,there is a flaw in the proof. Hessian(x0) being non-degenerate does not imply f is P.D or N.D hence x(0) not necessarily a local extremum. e.g saddle point.
Here is my version: for x in N(0), the Hessian(x) is non-degenerate since eigenvalues of matrix continously depends on entry hence x.
in N(0) any x,y we have
grad(x)-grad(y)=H(x-y)  +o(|x-y|^2)
H is non-degenrate ,then the map from x to grad(x) is one-to one linear map in N(0).therefore grad(x) implies x=0