Show that a surface of revolution locally admits a conformal parametrization

Question

Show that a surface of revolution locally admits a conformal parametrization, i.e. $E = G$, and $F = 0$.
I can calculate that for a surface general surface of revolution i.e. $$f(u,v)=(r(v)\cos u,r(v)\sin u,h(v))$$ we have that \begin{align*} f_u&=(-r(v)\sin u,r(u)\cos u,0)\\ f_v&=(r^\prime(v)\cos u,r^\prime(v)\cos u,h^\prime(v))\\ E&=f_u\cdot f_u=r^2(v)\\ F&=f_u\cdot f_v=0\\ G&=f_v\cdot f_v=r^{\prime2}(v)+h^{\prime2}(v) \end{align*} then I don not know how to find the conformal parameterization, because I don't know how to make $$r^2(v)=r^{\prime2}(v)+h^{\prime2}(v).$$ Maybe I did not understand the meaning of "local" correctly?
I tried to follows Ted's comment so that assume $$v=\phi(\bar{v})$$ then \begin{align*} \bar{E}&=r^{2}(\phi(\bar{v}))\\ \bar{G}&=r^{\prime 2}(\phi(\bar{v}))\phi^{\prime2}(\bar{v})+h^{\prime 2}(\phi(\bar{v}))\phi^{\prime2}(\bar{v})=\phi^{\prime2}(\bar{v}) \end{align*} let $\bar{E}=\bar{G}$ so $$\phi^{\prime2}(\bar{v})=r^{2}(\phi(\bar{v})) \Rightarrow \phi^{\prime}(\bar{v})=r(\phi(\bar{v}))$$