This problem is from Wolfgang Kühnel. The original statement is "Construct a non-planar C∞-curve which is a Frenet curve except for a single point, and outside of this point satisfies τ ≡ 0". I've been constructing some functions yet I haven't had any luck yet. In my search for a solution I ran into the following problem.
If the torsion of the curve is zero at every point (expect possibly in the point where it isn't a Frenet curve) then we have that $det(\alpha',\alpha'',\alpha''')=0$, which is given by the torsion formula $\tau=\frac{det(\alpha',\alpha'',\alpha''')}{|\alpha'\times \alpha''|}$. For the curve to be a Frenet curve the vectors $\alpha', \alpha'',\alpha'''$ must be linearly independent, but the determinant of the matrix generated by said vectors is zero, which tells us that the solution of a linear equation system given byt said vectors in not unique, thus making them linearly dependent.
With this argument I think that it is impossible to construct said curve. It also sounds like a constradiction since the fact that the curve is non-planar, yet the torsion is zero (I've always thought that $\tau=0\implies$ planar). Any input or hints for the problem is appreciated!
I will show that the union of two curves below satisfies the requirement. $$ C_1: \begin{cases} y=e^{-\frac{1}{x^2}}, & x\geq 0, \\ z=0, & \\ \end{cases} $$ $$ C_2: \begin{cases} z=e^{-\frac{1}{x^2}}, & x\geq 0, \\ y=0. & \\ \end{cases} $$ Clearly $C_1$ and $C_2$ are all $C^{\infty}$ except on $(0, 0, 0)$. Note that the derivations at $(0, 0, 0)$ are all $(1, 0, 0)$ on two curves, hence $C=C_1\cup C_2$ is $C^{\infty}$. Both $C_1$ and $C_2$ are planar curves, so $\tau=0$ for every point except $(0,0,0)$. It is trivial that $C=C_1\cup C_2$ is not in any plane, and $C$ is not a Frenet curve on $(0,0,0)$.