A bit of trivia (can't think of any use of it now)

The statement follows from considering the conformal change of curvature formula in two dimensions. By a rescaling, we can assume $e^{2u}g$ has Gauss curvature -1. Then $u$ must solve \[\triangle_g u - e^{2u} = K_g\] where $\triangle_g, K_g$ are the Laplace-Beltrami operator and Gauss curvature associated with $g$. Now hit the equation with the Lie derivative $\mathcal{L}_V$. Since $V$ is Killing relative to $g$, its Lie derivative commutes with the Laplace-Beltrami operator, and kills $K_g$. So we have \[\triangle_g V(u) - 2 e^{2u}V(u) = 0. \] Now we multiply by $V(u)$ throughout, and integrate over $M$, we have that \[ \int_M |\nabla V(u)|^2 + 2 e^{2u} V(u)^2 dvol_M = 0 \] which leads to our conclusion.

Observe that the same argument works for conformally flat Riemann surfaces. That is: let $(M,g)$ be a 2-dimensional compact Riemannian manifold with non-positive Euler characteristic, and let $g_0$ be the conformal metric with constant Gauss curvature. Then any infinitesimal symmetry of $g$ is a symmetry of $g_0$.

For surfaces of positive Euler characteristic (well, the topological 2-sphere), this argument doesn't work. And in fact, the statement is false. This is related to the fact that the 2-sphere has a really huge conformal group (any fractional-linear/Moebius transformation of the complex plane gives rise to one such) (which also causes a bit of technical difficulty when considering the uniformisation problem for the sphere). For example, under the stereographic projection of the 2-sphere to the complex plane $\mathbb{C}$, the canonical rotation vector field is given by $\frac{\partial}{\partial\theta}$ in polar coordinates. The metric can be written as $16dz d\bar{z}/(4 + |z|^2)^2 $. Now take the conformal transformation $w = (z+1)/(z+2)$, which implies that $dw = dz/(z+2)^2$. The pull-back metric $16dwd\bar{w} / (4 + |w|^2)^2$ differs from the standard metric by a factor of $[(4 + |z|^2) / (4|z+2|^2 + |z+1|^2)]^2$, which is emphatically not constant along integral curves of $\frac{\partial}{\partial\theta}$.

One final remark, by Bochner's theorem if $(M,g)$ already has negative curvature (but not everywhere constant), there cannot be a non-trivial Killing vector field. Hence the little bit of trivia is only useful the Gauss curvature of $g$ is positive somewhere.