$f \colon \mathbb R^n \to \mathbb R$ is differentiable in $x_0$ if there exists a functional $L$ $$f(x_0+h)-f(x_0)-Lh=o(|h|),$$ as $|h|\to 0.$ Here $o(|h|)$ denotes a function going to $0$ faster than $|h|$ depends on $x_0.$ Denote $Df(x_0)=L$.
$f$ is continuously differentiable in $\mathbb R^n$ if it is differentiable in $x_0$ for every $x_0$ and $Df(x)$ is continuous on $\mathbb R^n.$
What are the conditions guaranteing that $o$ is independent of $x_0$? Intuitively it should be that $Df(x)$ is uniformly continuous, is that the case?
Fix any point $x_0\in\mathbb R^n$ and any vector $e\in\mathbb R^n$ of length $1$. Define a function $g:\mathbb R\to \mathbb R$ putting $g(t)=f(x_0+te)$ for each $t\in\mathbb R$. I expect that for each $t\in\mathbb R$ the differentiability of $f$ at $x_0+te$ implies that $g'(t)=Df(x_0+te)e$. For each $t>0$ by Lagrange's theorem there exists $s(t)\in (0,t)$ such that $$f(x_0+te)-f(x_0)=g(t)-g(0)=g'(s(t))t=Df(x_0+s(t)e)te.$$ Then $$f(x_0+te)-f(x_0)-Df(x_0)te=(Df(x_0+s(t)e)-Df(x_0))te.$$
Suppose now that $Df(x)$ is uniformly continuous, that is for each $\varepsilon>0$ there exists $\delta(\varepsilon)>0$ such that for each $x,x'\in\mathbb R^n$ with $\|x-x'\|<\delta(\varepsilon)$ we have $\|Df(x)-Df(x')\|\le\varepsilon$, where the norm of a functional $C$ from $\mathbb R^n$ to $\mathbb R$ is defined as $\sup \{|Cy|:y\in\mathbb R^n\mbox{ and }\|y\|=1\}$.
Given any $\varepsilon>0$, pick any $t\in (0,\delta(\varepsilon))$. Then $\|(x_0+s(t)e)-x_0\|=s(t)\in (0,\delta(\varepsilon))$, so $\|Df(x_0+s(t)e)-Df(x_0)\|\le \varepsilon$, and so $|(Df(x_0+s(t)e)-Df(x_0))te|\le t\varepsilon$. Then $|f(x_0+te)-f(x_0)-Df(x_0)te|\le t\varepsilon$. This should imply the required conclusion.