I have found a proof but I'd like to see if it's correct.
Consider a continuous function $f:\mathbb{R}\to\mathbb{R}$ that is differentiable on $\mathbb{R}_0$. Suppose that $\lim_{x\to 0}f'(x)$ exists and is finite. Then $f$ is differentiable in $0$ as well.
$\textbf{Proof}:$ Let $\varepsilon > 0$. Then $f$ is continuous on $[0,\varepsilon]$. The function is also differentiable on $(0,\varepsilon)$. Hence by the mean value theorem there is a $\xi\in(0,\varepsilon)$ such that $$f'(\xi) = \dfrac{f(\varepsilon)-f(0)}{\varepsilon-0}.$$ Now, notice that $$\lim_{\varepsilon \to 0}f'(\xi)=\lim_{\xi \to 0}f'(\xi) < \infty$$ and hence $$\lim_{\xi \to 0}f'(\xi) = \lim_{\varepsilon \to 0}\dfrac{f(\varepsilon)-f(0)}{\varepsilon-0}<\infty.$$ So $f$ is differentiable in 0.