Let $u$ be a smooth real valued function on $\mathbb{R}^2$. Is it possible for the derivative $\frac{\partial u}{\partial y}$ to have an isolated zero? I think the answer is no. Here is my reasoning:
Let $x\in \mathbb{R}^2$. If $x$ is a critical point of $u$, then I think by the Poincare index theorem there must exist a point in every ball around $x$ such that $\frac{\partial u}{\partial y}=0$ so $x$ is not an isolated critical point of $\frac{\partial u}{\partial y}$. Now if $x$ is not a critical point of $u$, then locally around $x$, the level set of $u$ looks like a bunch of parallel curves "foliating" the ball. Then there should be a point in every ball around $x$ such that $\frac{\partial u}{\partial y}=0$ which means that $x$ is not an isolated critical point of $\frac{\partial u}{\partial y}$.
Is this correct? Or is there some obvious counterexample I'm not seeing?
My original question was to determine whether or not the directional derivative (in a particular direction) of a smooth function can have isolated critical points, which I believe is equivalent to the above.
Any help would be appreciated thank you.