Consider a smooth surface $S$ in $\mathbb{E}^3$ (It may not be complete).
Here a unit speed smooth curve $c:[0,l]\rightarrow S\subset \mathbb{E}^3$ on $S$ is an asymptotic line if the second derivative $c''(t)$ is in the tangent space $T_{c(t)}S$.
Introduction : Is there a surface $S$ s.t. there is a closed curve $c$ i.e. $c(0)=c(l)$ s.t. $c$ is an asymptotic line and a Gaussian curvature at a point $c(t)$ is negative for all $0\leq t\leq l$ ?
(1) Consider a torus of revolution $f(s,t)=((R+r\cos\ t)\cos\ s,(R+r\cos\ t)\sin\ s,r\sin\ t)$. Here $c (s)=(R\cos\ s,R\sin\ s,r)$ is an asymptotic line. Further, note that Gaussian curvatures along the curve $c$ are zero.
(2) Consider $z=f(x,y)=x^2-y^2$. The graph surface has a negative Gaussian curvature at all points. Further, for any constant $C$, $\{(t,t+C,-2tC-C^2)|t\in \mathbb{R}\}$ is an asymptotic line, which is not closed.
When $c(t)=(\overline{c}(t),f\circ\overline{c} (t))$ has a unit speed, then consider $c''(t)$. So $c$ is an asymptotic line when $ \langle\overline{c}'(t),{\rm Hess}\ f\cdot \overline{c}'(t)\rangle=0$ where $\cdot $ is a matrix multiplication and $\langle\ ,\ \rangle$ is an inner product.
Consider a set $A = \{(x,y)| |x|+|y|=1\}$ homeomorphic to $\mathbb{S}^1$. Then $\{(a,f(a)) | a\in A\} $ is union of $4$ asymptotic lines.
Question : Assume that $S$ is a graph surface i.e. $z=f(x,y)$. When $c$ is unit speed curve on $S$ and $\overline{c}$ is its orthogonal projection into $xy$-plane, then assume that $c$ is a closed asymptotic line s.t. Gaussian curvature along the whole $c$ is negative. Then $\overline{c}$ does not enclose a star-shaped domain.
Reference : 27p. in the book Puzzles in geometry which I know and love - Petrunin