Showing a Critical Point is Isolated

Question

I'm looking at the critical point $(0,0)$ of the function $f(x,y)=x^2+4y^3$. Since the Hessian is singular, we cannot conclude anything about the local extrema. However, a graph of this function clearly shows that $(0,0)$ is neither a local max nor min. I'm wondering if there is a way to go about showing this without just referring to the graph. Further, is there a way to show that $(0,0)$ is isolated or not isolated?

Answer 1

A critical point $p$ is not a local extremum if every neighborhood of $p$ contains a point $q$ with $f(q) > f(p)$ and a point $q'$ with $f(q') < f(p)$.

In this case, $f(0,0) = 0$, but for any $\epsilon>0$, $f(\epsilon,0) = \epsilon^2 > 0$, and $f(0,-\epsilon) = -4\epsilon^3 < 0$.