$S^2$ is the topological sphere. Consider two differet metric on it, respectively denote $(S^2,g_1), (S^2, g_2)$. The two metric are conformally equivalent if $$ g_1=\lambda g_2 $$ where $\lambda:S^2\rightarrow (0,+\infty)$. Seemly, the Uniformization theorem imply that $(S^2,g_1)$ and $(S^2, g_2)$ are conformally equivalent. Therefore, I try some specific example.
The unit sphere in 3-dim Euclidean space \begin{align} &r(\theta,\varphi)=(\sin\theta\cos\varphi, \sin\theta\sin\varphi,\cos\theta) \\ &r_\theta=(\cos\theta\cos\varphi,\cos\theta\sin\varphi,-\sin\theta) \\ &r_\varphi=(-\sin\theta\sin\varphi, \sin\theta\cos\varphi, 0) \end{align} $$ g_1=\left( \begin{array}{1} 1 &0 \\ 0 &\sin^2\theta \end{array} \right) $$ The ellipsoid in 3-dim Euclidean space \begin{align} &r(\theta,\varphi)=(\sin\theta\cos\varphi, \sin\theta\sin\varphi, 2\cos\theta) \\ &r_\theta=(\cos\theta\cos\varphi,\cos\theta\sin\varphi,-2\sin\theta) \\ &r_\varphi=(-\sin\theta\sin\varphi, \sin\theta\cos\varphi, 0) \end{align} $$ g_2=\left( \begin{array}{1} 1+3\sin^2\theta &0 \\ 0 &\sin^2\theta \end{array} \right) $$ Obviously, when $\sin\theta \ne 0$, there is not $\lambda$ such that $g_1=\lambda g_2$. How to understand it ?