The two surfaces $S_1$, $S_2$ parameterised respectively by
$$\sigma_1(u,v) = (u\cos v,u\sin v, \ln u)$$ $$\sigma_2(u,v) = (u\cos v, u\sin v, v),$$
are, as I understand, not locally isometric. How can this be proven? They have the same Gaussian curvature, so that's not enough. The coefficients of the first fundamental form in each are not equal, but is that enough to conclude that they're not isometric? I've been taught that the coefficients of the first fundamental form depend on the parameterisation, so just because those of the above patches don't agree, how can we be sure that there isn't some local isometry $f$ such that $f\circ \sigma_1$ is a parameterisation of $S_2$?
Since the curvature functions are functions of $u$ and must agree at corresponding points, any reparametrization must leave $u$ as is. This means that, no matter how you alter your $v$ parameter, the first entries of the first fundamental form cannot change, and they are most definitely not equal.