Let $f \in C^{0,1}(\mathbb R^2)$, that is $f$ is continuous and the second partial derivative $\partial f/\partial x$ exists and is continuous. Then, by Taylor's formula, \begin{equation*} R(t,x,h) = \frac{f(t,x+h) - f(t,x)}{h} - \frac{\partial f}{\partial x}(t,x) \end{equation*} defines a function in $(t,x,h) \in \mathbb R^3$, such that, for all $(t,x) \in\mathbb R^2$, $R(t,x,\cdot)$ is a continuous function taking the value zero in zero.
If $f \in C^1(\mathbb R^2)$, then $R$ is continuous as a function $\mathbb R^3 \rightarrow \mathbb R$.
Question. For general $f \in C^{0,1}(\mathbb R^2)$, is $R$ continuous as a function $\mathbb R^3 \rightarrow \mathbb R$? If not, can you give a counterexample?
This is not true.
Counterexample: $f(t,x) = \frac{2tx}{\sqrt{t^2+x^2}}$, the typical example of a componentwise differentiable, but not differentiable function does the job, since in that case, \begin{equation*} R(h,h,h) = \frac{8 - 3 \sqrt{10}}{\sqrt{20}} \neq 0 \end{equation*} for $h > 0$.