Assume that $f(x)$ has a continuous derivative in some open interval $O$ containing $x_0$. Then for sufficiently small $h$ we may apply the mean value theorem to the interval $[x_0-h,x_0+2h]$ and conclude that there is a real number $c_h$ with $x_0-h for which $ \frac{f(x_0+2h)-f(x_0-h)}{(x_0+2h)-(x_0-h)}=f'(c_h).$ Then since $c_h$ approaches $x_0$ as $h\to 0$ (it's trapped between two things which both approach $x_0$), and using continuity of $f'(x)$, we see that as $h \to 0$ the difference quotient $ \frac{f(x_0+2h)-f(x_0-h)}{3h} \to f'(x_0).$
The statement of the question might not hold for functions $f$ which are only assumed to be differentiable at $x_0$, but I can't think of an example of this immediately.
Re-statement of a previous approach: break three times the given difference quotient into three parts: $\frac{f(x_0+2h)-f(x_0+h)}{h}+\frac{f(x_0+h)-f(x_0)}{h}+\frac{f(x_0)-f(x_0-h)}{h}$ and note that each of these difference quotients approaches $f'(x_0)$ as $h \to 0$, under the assumption that $f'(x)$ is continuous in an open interval about $x_0$. This may be in a way more intuitive to some than the above version using the mean value theorem.
EDIT: I just noticed that this "previous approach" shows one only need assume that $f(x)$ is differentiable at $x_0$. To see this, note that the second and third terms approach respectively the right hand and left hand derivatives of $f(x)$ at $x_0$. And note that the first term may be broken into $2\frac{f(x_0+2h)-f(x_0)}{2h} + \frac{f(x_0)-f(x_0+h)}{h}.$ As $h \to 0$ first term here approches $2f'(0)$ and the second approaches $-f'(x_0)$. So combining all three terms, or all four terms given the splitting of the earlier first term into two parts, we get a net of $2f'(x_0)-f'(x_0)+f'(x_0)+f'(x_0)=3f'(x_0)$ for the limit, which as before on division by 3 gives the desired $f'(x_0).$ So no need of any extra assumptions, beyond that $f'(x_0)$ exists.