Hello, Kaitosan!
Can piecewise functions be differentiable throughout? They certainly can . . . We can easily create one.
The graph must be continuous throughout and have equal derivatives at the "junctions".
Consider the graph of the parabola $\displaystyle y \:=\:x^2$
Bisect it through its axis of symmetry.
Move each "half" one unit to the left and right, respectively.
The graph looks like this: Code:

*  *

*  *
*  *
*  *
      *+*      
1  1
And we have: .$\displaystyle f(x) \;=\;\begin{Bmatrix} (x+1)^2 & & x \leq 1 \\ 0 & & 1 < x < 1 \\ (x\:{\color{red}}\:1)^2 & & x \geq 1 \end{Bmatrix}$
This is the example I gave when asked this very question
. . on an exam while in college . . . centuries ago.
Yes . . . a silly typo . . . Thanks for the headsup.