Suppose that we have a differentiable function $f\colon I\to\mathbb{R}$ on some interval $I\subset\mathbb{R}$. The difference quotient of $f$ is defined by $$\Delta_{f}\colon I\times I\to\mathbb{R},\qquad\Delta_{f}(x,y):=\begin{cases}\dfrac{f(x)-f(y)}{x-y} & x\not=y \\ & \\ f'(x)=f'(y) & x=y\end{cases}.$$ It is clear that $\Delta_{f}$ is separately continuous by definition of differentiability.
(*) According to this webpage, $\Delta_{f}$ is jointly continuous if and only if $f'$ is continuous (i.e. $f$ is C${}^{1}$).
Now assume that $f$ is twice differentiable and define the second order difference quotient of $f$ by $$\Delta_{f}^{2}\colon I\times I\to\mathbb{R},\qquad\Delta_{f}^{2}(x,y):=\begin{cases}\dfrac{f'(x)-\frac{f(x)-f(y)}{x-y}}{x-y} & x\not=y \\ & \\ \tfrac{1}{2}f''(x)=\tfrac{1}{2}f''(y) & x=y\end{cases},$$ where the factor $\tfrac{1}{2}$ is "justified" by Taylor's theorem. I was wondering if (*) generalizes as follows.
(**) Is it true that $\Delta_{f}^{2}$ is jointly continuous if and only if $f''$ is continuous (i.e. $f$ is C${}^{2}$)?
In a similar fashion one can define $\Delta_{f}^{k}$ and ask whether $\Delta_{f}^{k}$ is jointly continuous if and only if $f$ is C${}^{k}$. However, if the $k=2$ case is true, then I suspect that the general case can be proven inductively.