Does there exist a continuous function $f:\mathbb{R}\to\mathbb{R}$ so that $f$ is differentiable exactly at one point?
Yes. Idea: choose a function which is continuous but nowhere differentiable. Then multiply it with $x^2$, say.
$$f(x) = \begin{cases} x^2 & x \in \Bbb{Q} \ -x^2 & x \in \Bbb{R}\setminus\Bbb{Q} \end{cases}.$$
$$f(x) = \begin{cases} x^2 & x \in \Bbb{Q} \ -x^2 & x \in \Bbb{R}\setminus\Bbb{Q} \end{cases}.$$
– Sangchul Lee Aug 26 '12 at 08:58