Given $U = (0,1)\times(0,\pi)$ and $P: U \to \mathbb{R}^3$ define as $$P(u,v) = (u\cos v, u\sin v, \phi(v))$$ where $\phi(v)$ is a bijective function $C^\infty(0,\pi)$. Given the curves $$ \gamma_1(t)=(\cos t,\sin t,\phi(t))$$ $$\gamma_2(t) = \bigg({\sqrt{2}\over{2}}t, {\sqrt{2}\over{2}}t, \phi\bigg({\pi\over{2}}\bigg)\bigg) $$ say if they're curvature lines for $P(U)$.
I'm a bit unsure on what i've done, I've calculated first and second fundamental form, and then I've computed the Weingarten matrix. Then I notice that with $u=1, v=t$
$$\gamma_1'(t)= 0\cdot P_u|_{(1,t)}+1\cdot P_v|_{(1,t)}$$
So denoted with $S$ the Weingarten matrix
$$S\cdot(0,1)^T = \bigg({-\phi'(t)\over(1+\phi'(t)^2)^{1/2}},{\phi''(t)\over(1+\phi'(t)^2)^{3/2}}\bigg)^T $$
And since the obtained vector isn't a multiple of $(0,1)^T$, $\gamma_1$ isn't a curvature line. For $\gamma_2$ i notice that there aren't values of $u,v$ such that $P(u,v) = \gamma_2(t)$, so $\gamma_2\notin P(U)$ and it can't be a curvature line for $P(U)$.
Does anything I said makes sense? Is this the only method to verify if a line is a curvature line for some surface in $\mathbb{R}^3?$ Thank you very much.