Is there a function that has point reflection in $(0,0)$ and that is only not differentiable in $x=0$? 
Is there a function that has point reflection in $(0,0)$ and that is
  only not differentiable in $x=0$?

I don't think that this is possible, but I can't prove it - can you find an example where this true?
 A: Hint:
If by "has point reflection in $(0,0)$" you mean that the function is odd, consider $f : \mathbb{R} \to \mathbb{R}$ given by $$f(x) = (\operatorname{sign} x)\sqrt{|x|}$$
Then for every $x \ne 0$ we have $f'(x) = \frac1{2\sqrt{|x|}}$, but $f$ is not differentiable at $0$.
A: Yes, such functions exist and are easy to make. Just take any continuous function $f:[0,2]\to\Bbb R$ with $f(0)=0$, and which is differentiable on the interior of the interval, but not on the right in$~0$; then complete it to an odd function by setting $f(-x)=-f(x)$ for $x\in[0,2]$. Right-differentiability in the origin can be frustrated either by the graph of $f$ going for instance above any line $y=ax$ for $a\in\Bbb R$, near the origin, or by visiting more than one such line infinitely often near the origin. An example of the first phenomenon is $f(x)=\sqrt x$ (which made odd becomes $f(x)=\operatorname{sgn}(x)\sqrt{|x|}$), and of the second phenomenon an example is $f(x)=x\cos(\frac1x)$ for $x\neq0$ with $f(0)=0$ (which already defines an odd function).
