Let $f$ be a real function and let $f'$ be its strong derivative, i.e., for all $x\in\mathbb R$
$$f'(x)=\lim_{x_1,x_2\to x}\frac{f(x_2)-f(x_1)}{x_2-x_1},$$
where $\lim_{x_1,x_2\to x}\frac{f(x_2)-f(x_1)}{x_2-x_1}$ is the $a\in\mathbb R$ such that for every $\epsilon>0$ there is a $\delta>0$ such that $|\frac{f(x_2)-f(x_1)}{x_2-x_1}-a|<\epsilon$ whenever $x_1\neq x_2$ and $\max\{|x-x_1|,|x-x_2|\}<\delta$.
Assume that both $f$ and $f'$ are defined on all of $\mathbb R$ and are both uniformly continuous. I need help to prove that the function $F:\mathbb R^2\to\mathbb R$ defined by
$$F(x_1,x_2)=\begin{cases} \frac{f(x_2)-f(x_1)}{x_2-x_1} & \text{if }x_1\neq x_2 \\ f'(x_1) & \text{if }x_1=x_2 \end{cases}$$
is also uniformly continuous. And I need that proof to not make use of any theorems about derivatives.
Background: Some time ago I asked the same question about the standard derivative: Uniform continuity of difference function. Because of the lack of answers, I'm beginning to suspect that it cannot be done. I'm hoping that strengthening the assumption will do the trick.