Suppose that $x_{0}$ is a local maximum (a similar proof applies if $x_{0}$ is a local minimum). Then there exists $\delta>0$ such that $(x_{0}-\delta,x_{0}+\delta)\subset(a,b)$ and such that we have $f(x_{0})\geq f(x)$ for all $x$ with $|x-x_{0}|<\delta$. Hence for $h\in(0,\delta)$ we notice that it holds

Since the limit of this ratio as $h\to 0^{+}$ exists and is equal to $f^{{\prime}}(x_{0})$ we conclude that $f^{{\prime}}(x_{0})\leq 0$. On the other hand for $h\in(-\delta,0)$ we notice that

but again the limit as $h\to 0^{+}$ exists and is equal to $f^{{\prime}}(x_{0})$ so we also have $f^{{\prime}}(x_{0})\geq 0$.

Hence we conclude that $f^{{\prime}}(x_{0})=0$.

To prove the second part of the statement (when $x_{0}$ is equal to $a$ or $b$), just notice that in such points we have only one of the two estimates written above.