Let $f:[0,1]\times [0,1]\to \mathbf{R}$ such that $f$ is infinitely differentiable and always non-negative. Additionally, $f$ only has finitely many zeros. WLOG assume that $f(a,b) = 0$ for some $(a,b) \in [0,1]\times [0,1]$. Suppose that $\frac{\partial}{\partial x}f(a,b) = 0$, $\frac{\partial}{\partial \xi}f(a,b) = 0$, $\frac{\partial^2}{\partial x\partial \xi}f(a,b) = 0$, and $\frac{\partial^2}{\partial x^2}f(a,b) = 0$. However, $\frac{\partial^2}{\partial \xi^2}f(a,b) \ne 0$. In other words, the first nonzero derivative is the second partial derivative with respect to $\xi$. Let $B_{\delta}$ be an open ball centered about the point $(a,b)$ of some radius $\delta >0$. Do there exist constants $c, k$ (dependent on $\delta$) such that $$ |f(x,\xi)| \geq c ((x-a)^2 + (\xi-b)^2)^{\frac{k}{2}} $$ for almost every $(x,\xi) \in B_{\delta}$?
My guess is yes with $k = 2$ as that is the order of the first nonzero derivative. However, I'm not quite able to show this. I also believe that this can be extended in generality. Any advice would be appreciated.
The answer is clearly no.
Take for example $f(x, \xi) = e^{-\frac{1}{(x-a)^2}} + (\xi-b)^2$ for $x \neq a$ and $f(a,\xi) = (\xi - b)^2$. This works because for any $n \ge 1$ $\frac{\partial^n f}{\partial x^n}(a,\xi)=0$.