I want to show that a geodesic with nowhere vanishing curvature is a plane curve if and only if it is a line of curvature.
I have done the following:
Let $\gamma$ be a geodesic with nowhere vanishing curvature and a plane curve.
Let $N$ be the unit normal of the surface.
We have that $\|N\|^2=1 \Rightarrow N\cdot N=1 \Rightarrow (N\cdot N)'=0\Rightarrow N'\cdot N=0 \Rightarrow N'\bot N$.
$\gamma'$ is the tangent vector of $\gamma$.
Since $N'$ is perpendicular to $N$, which is perpendicular to the curve $\gamma$ we have that $N'$ is parallel to $\gamma$, so it is also parallel to $\gamma'$. So $N'=-\lambda \gamma'$, for some $\lambda$, which means that $\gamma$ is a line of curvature.
Is this correct?
For the other direction:
Let $\gamma$ be a geodesic with noehere curvature and a line of curvature.
Without loss of generality, we suppose that $\gamma$ is unit-speed.
Let $a=N\times\gamma$. Differentiating this we get $a'=N'\times\gamma'+N\times\gamma''$. Since $\gamma$ is a line of curvature we have that there is a $\lambda$ such that $N'= \lambda\gamma'$, i.e., $N'$ is paralle to $\gamma'$. So $N'\times\gamma'=0$. Since $\gamma$ is geodesic we have that $N\times\gamma''=0$.
How could we continue to conclude that the geodesic is a line of curvature.