Are all continuous one one functions differentiable?

Not by a long shot. Take, for example, the function

$$f(x) = \begin{cases}x & x\leq 0\\ 2x & x\geq 0\end{cases}$$

Which is continuous and one-to-one on $\mathbb R$, but is not differentiable at $0$.

This is of course just one example, but in general, any time you "stick" two functions together at a point where their derivatives are not equal, like in my example, you can cause the resulting function to have a point at which it is not differentiable.

$x^{1/3}$ is not differentiable at $0$. See its graph above. It's qualitatively different from the example given by 5xum.

The Cantor function $ +\, x$ is an example of a function that's continuous and one-to-one, but non-differentiable at uncountably many points.

There's a limit to how bad an example can get. The set of points where a continuous one-to-one functions is non-differentiable always has Lebesgue measure $0$.

Well, you could do something like: Let $a_i$ be an enumeration of the rationals.

Define $f_i(x)$ as a continuous, nondecreasing function, strictly between $0$ and $1$ which is differentiable everywhere but $a_i$.

Define $g(x) := x + \sum_{i=0}^{\infty} f_i(x)\times 2^{-i}$

That should give a function that is continuous, but not differentiable at any rational number. (It's differentiable at every irrational number, though.)