I am having trouble proving the following statement, which at first seemed intuitively true to me.
Let $S$ be a surface in $\mathbb{R}^{3}$. Suppose that there is a curve $\gamma$ in $S$ whose all points are planar, i.e., the second fundamental form $\alpha$ (or, equivalently, the shape operator) of $S$ vanishes at all points of $\gamma$. Does this imply that $\gamma$ is part of a straight line?
This question is related to the existence of non-straight asymptotic curves in $S$. It is well-known that a curve $\gamma$ such that $\alpha(\dot{\gamma},\dot{\gamma})=0$ need not be part of a straight line.
EDIT: As pointed out by Arctic Char, the claim is not true in general. What happens if we assume that, for every open neighborhood $U$ (in $S$) of the curve $\gamma$, there is no plane $P$ such that $U \subset P$?
Take for example the surface parametrised by $$ x(u,v) = (u, v, (u^2-v)^3). $$ It easy to see that the second derivatives $x_{uu}$, $x_{uv}$, $x_{vv}$ are zero along the curve $v = u^2$, and hence the shape operator vanishes along this curve. However the curve is a parabola in the $xy$-plane, not a straight line.