$\newcommand{\R}{\mathbb{R}} \newcommand{\al}{\alpha}$ This is a partial answer: Assuming additionally that $D^\al f$ exists in $C(\R^n)$ for $|\al|=1$, let us show that $D^\al f$ is bounded. Indeed, without loss of generality $\al=e:=(1,0,\dots,0)\in\R^n$. We have $|f|\le M$ and $|D^{2e}f|\le M_2$ for some real $M,M_2\ge0$.
By fixing arbitrary values of the last $n-1$ arguments of the functions, we reduce the consideration to the case $n=1$, so that $D^e f=f'$, the derivative of $f$. Then, by an appropriate version of Taylor's theorem, for any real $x$ and some $c_\pm=c_\pm(x)\in[-1,1]$ we have \begin{equation} f(x\pm1)=f(x)\pm f'(x)+c_\pm M_2/2, \end{equation} whence, by subtraction, \begin{equation} |f'(x)|\le\tfrac12|f(x+1)-f(x-1)|+M_2/2\le M+M_2/2, \end{equation} so that $|f'|\le M+M_2/2$.