I have two surfaces, with parametrizations:
$$\mathbb{x}(u,v)=(u\cos v,u\sin v,\log u) \\ \mathbb{\bar{x}}(u,v)=(u\cos v,u\sin v, v).$$
I need to proof that the map $\mathbb{\bar{x}}\circ \mathbb{x}^{-1}$ is not an isometry, but I don't know how to find $\mathbb{x}^{-1}$; Any hints?
Basically, I need to find a function $\mathbb{x}^{-1}:\mathbb{R}^{3}\rightarrow\mathbb{R}^{2}$ such that $\mathbb{x}^{-1}(u\cos v, u\sin v, \log u)=(u,v).$
Remember that a diffeomorphism $\varphi: S \to \overline{S}$ between surfaces is an isometry when for all $p \in S$ and all pairs $w_1, w_2 \in T_pS$ we have:
$$ \left\langle w_{1}, w_{2}\right\rangle_{p}=\left\langle\mathrm{d} \varphi_{\mathrm{p}}\left(w_{1}\right), \mathrm{d} \varphi_{p}\left(w_{2}\right)\right\rangle_{\varphi(p)} $$
By the polarization identity, there is no loss of generality in assuming $w_1 = w_2$. Also, it's easy to show that this happens iff $\varphi$ preserves the first fundamental form (which happens iff $\varphi$ preserves lengths of curves). I shall prove that this does not happen for the diffeomorphism $\varphi = \bar{x} \circ x^{-1}: x(U) \subset S \to \overline{x}(U) \subset \overline{S}$ described here (without explictly calculating $x^{-1}$!). Indeed, let $\alpha: (-\varepsilon, \varepsilon) \to S$ be a smooth curve in $S$ such that $\alpha(0) = p$ and $\alpha'(0) = w \in T_pS$, and let's write $\alpha(t) = x(u(t), v(t))$. We have:
$$\langle w, w \rangle_p = \mathrm{I}_p(w) = E(u')^2 + 2Fu'v' + G(v')^2$$
Now, let $\overline{\alpha} = \varphi \circ \alpha : (-\varepsilon, \varepsilon) \to \overline{S}$ be the image of $\alpha$ by $\varphi$. It's clear that $\overline{\alpha}(0) = \varphi(p)$ and that $\overline{\alpha}'(0) = \mathrm{d}\varphi_p(w)$ (chain rule in action here). Furthermore, $$\overline{\alpha}(t) = \varphi(\alpha(t)) = \overline{x}(u(t), v(t))$$
Therefore
$$\begin{aligned} \langle \mathrm{d} \varphi_p(w), \mathrm{d} \varphi_p(w) \rangle_{\varphi(p)} &= \overline{\mathrm{I}}_{\varphi(p)}(\mathrm{d} \varphi_p(w)) \\ &= \overline{E}(u')^2 + 2 \overline{F} u' v' + \overline{G}(v')^2 \end{aligned}$$
Now, an easy calculation shows that $$E = 1 + \frac{1}{u^2}, F = \overline{F} = 0, G = u^2, \overline{E} = 1, \overline{G} = 1 + u^2$$.
So we have
$$\begin{aligned} &\langle \mathrm{d} \varphi_p(w), \mathrm{d} \varphi_p(w) \rangle_{\varphi(p)} = (u')^2 + (v')^2 + u^2 (v')^2 \\ &\langle w, w \rangle_p = (u')^2 + \left( \frac{u'}{u} \right)^2 + u^2 (v')^2 \end{aligned}$$
Therefore, if $\varphi$ were an isometry, we would then have:
$$\begin{aligned} &\langle \mathrm{d} \varphi_p(w), \mathrm{d} \varphi_p(w) \rangle_{\varphi(p)} - (v')^2 = \langle w, w \rangle_p -\left( \frac{u'}{u} \right)^2 \\ &\left\langle w_{1}, w_{2}\right\rangle_{p}=\left\langle\mathrm{d} \varphi_{\mathrm{p}}\left(w_{1}\right), \mathrm{d} \varphi_{p}\left(w_{2}\right)\right\rangle_{\varphi(p)}\end{aligned} $$
which would imply $(v')^2 = \left( \frac{u'}{u} \right)^2 $ for any smooth curve of $S$ given by $\alpha(t) = x(u(t), v(t))$. This is false, because the equality above only happens when $v = \pm\ln(u) + C$, and $\alpha$ was chosen arbitrarily. This is a classical counter example proving that the converse of the theorema Egregium is false (this can happen when the gaussian curvature is not constant, which you can easily verify is the case here - when it happens that both gaussian curvatures are equal and constant, then by Minding's theorem both surfaces are locally isometric).
Note: if you really want to know the inverse, in this case it can be explicitly calculated (under a suitable domain for $u, v$, of course). It's given by $x^{-1}(x, y, z) = \left(\sqrt{x^2 + y^2}, \arctan\left(\frac{y}{x}\right) \right)$.