Suppose we have a plane curve $C: I \rightarrow \mathbb{R}^2$ that is continuously differentiable ($C^1$) and parameterized by arc length. Tangent lines exists at each point on the curve, so an "osculating circle" definition of curvature can be applied as follows. Given any $t_0 \in I$, find the tangent to the curve at $C(t_0)$. For any other $t \in I$, either 1) $C(t)$ is on the tangent line or 2) there is a unique circle containing $C(t_0)$ and $C(t)$ which is tangent to the curve at $t_0$. In the first case let $\kappa(t_0, t) = 0$ and in the second case let $\kappa(t_0,t) = 1/R$ where $R$ is the radius of the circle so defined. Finally let $$\kappa(t_0) = \lim_{t \rightarrow t_0} \kappa(t_0, t)$$ if the limit exists.
There is a second way to define curvature, in terms of the second derivitive. But $C$ might not have a second derivative at some points. Nevertheless, I believe that it will be twice differentiable if the curvature (defined in terms of osculating circles) exists at every point and is continuous.
I came to this conclusion after trying to make a counter example. I started with a simple example, an inverted semicircle: $C(t) = (x(t),y(t)) = (R \cos(t), R-R \sin(t))$ for $t \in [0,\pi]$. The point $t_0 =\pi /2$ is at the bottom of the semicircle, which happens to be at $(0,0)$. The curvature there, calculated via the osculating circle, reduces to: $$\lim_{t \rightarrow t_0} \dfrac{2 y(t)}{x(t)^2 + y(t)^2} = \dfrac{2(R-R\sin(t))}{2R^2 - 2R^2 \sin(t)} = \dfrac{1}{R}$$.
Starting with this curve, I tried to add "bumps" that got narrower and narrower as one approached $t= 0$. However, if the height of the bumps went to zero sufficiently fast that that $\kappa(t)$ was continuous at $t=0$, then the second derivative also existed at $t=0$.
Am I right or wrong, that continuity of $\kappa(t)$ guarantees that $C$ can be twice differentiated?