Mixed partials on the diagonal

Question

Let $I\subseteq\mathbb{R}$ be an open interval and $\varphi\in C^2(I^2;\mathbb{R})$ be twice continuously-differentiable on $I^2:=I\times I$.

We call $\varphi$ separable if it can be written $\varphi\equiv\varphi(x,y) = \alpha(x) + \beta(y)$ for some $\alpha,\beta: I\rightarrow\mathbb{R}$, and we call $\varphi$ non-separable if for each open $\mathcal{O}\subseteq I^2$ the restriction $\varphi_{\mathcal{O}}:=\left.\varphi\right|_{\mathcal{O}}$ is not separable.

Remark: For any $U\subseteq I^2$ open and connected, the function $\varphi_U$ is separable iff $\left.(\partial_x\partial_y\varphi)\right|_U = 0$; consequently $\varphi$ is non-separable if and only if the set $\{z\in I^2\,|\,\partial_x\partial_y\varphi(z)\neq 0\}$ is dense in $I^2$.

Question: Is it true that: If $\varphi$ is non-separable, then $\{x\in I \,|\,\partial_x\partial_y\varphi(x,x)\neq 0\}$ is dense in $I$?

Answer 1

No!

Example: Take $I=(0,1)$, and $$ \varphi(x,y)= \begin{cases} 0 & x=y\\ \exp(-1/(x-y)^2) & x<y\\ \exp(-2/(x-y)^2) & x>y. \end{cases} $$ By construction, $\varphi$ is smooth with derivatives of all order vanishing on the diagonal. But $\varphi$ isn't separable. Indeed, the mixed partial does not vanish anywhere else.