Let $S$ be a surface in $\mathbb{R}^{3}$ such that every regular curve $\gamma\subset S$ has nowhere vanishing curvature, that is $\kappa(z)\neq 0$ for all $z\in \gamma$. Does this imply that $S$ is a part of a sphere?



A surface on which all regular curves have nowhere vanishing curvature
I'm going to assume that you mean the following property



> 
> $S \subset \mathbb{R}^3$ has property (\*) if for any regular curve $\gamma\subset S$, the curvature vector $\vec{\kappa}$ of $\gamma$ as a curve in $\mathbb{R}^3$ is nowhere vanishing. 
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> 
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(if you mean the curvature of $\gamma$ as a curve in $S$, then no surface $S$ has such a property, as seen by any geodesic on $S$).


I claim that (\*) is equivalent to strict convexity of $S$. For any $\gamma \subset S$, (which we assume to be parametrized by unit speed, for simplicity) the curvature vector (in $\mathbb{R}^3$) is given by
$$
\vec{\kappa} = \nabla^{\mathbb{R^3}}\_{\dot\gamma}\dot\gamma = \nabla^{S}\_{\dot\gamma}\dot\gamma + h(\dot\gamma,\dot\gamma)\nu\_S,
$$
where $h$ is the second fundamental form of $S$ and $\nu\_S$ is the normal vector for $S$ in $\mathbb{R}^3$ (your conventions may vary, but it won't matter much anyways).


We know that $\nabla^{S}\_{\dot\gamma}\dot\gamma \in TS$. Hence, this is an orthogonal decomposition. Thus, for $\vec{\kappa} = 0$ it must be that $h(\dot\gamma,\dot\gamma) = 0$, which would contradict strict convexity.


On the other hand, if $S$ is not strictly convex, you can find a point $p$ and vector $V\in T\_pS$ with $h(V,V) = 0$. The geodesic through $p$ with velocity $V$ at $p$ will have vanishing $\mathbb{R}^3$ curvature at $p$.


