We'll consider $(N, g)$ a Riamannian Manifold and $\overline{g} = e^{2f}g$ a conformal metric. Let M be a hypersurface in N, $\overline{H}\_M$ and $H\_M$ the mean curvature of M with respect
to the metrics $\overline{g}$ and g, respectively. I would like some help to prove that


$$ \overline{H}\_M = e^{-f}( H\_M -2g( \nabla f, \eta)) $$


where $\nabla$ is the gradient with respect to metric g and $\eta$ is a normal vector field in M.



Relation between mean curvature and conformal metric
Let me use the transformation $\overline g = e^{2f}g$ to simplify some notations (and I guess your formula also use this convention). Near a point $p\in N,$ let $\{e\_i\}$ be an orthonormal frame with respect to $g,$ and $\eta$ be a normal. Then with respect to $\overline g,$ we have $\overline e\_i = e^{-f}e\_i$ form an orthonormal frame near $p$ and $\overline\eta=e^{-f}\eta$ being the normal. Then
\begin{align}
\overline h\_{ij}
& = \langle\overline\nabla\_{\overline e\_i}\overline e\_j,\overline\eta \rangle\_{\overline g}\\
& = e^{2f}\langle e^{-2f}\overline\nabla\_{e\_i}e\_j, e^{-f}\eta\rangle\_g\\
& = e^{-f}\langle \nabla\_{e\_i}e\_j-\delta\_{ij}\nabla f,\eta \rangle\_g\\
& = e^{-f}( h\_{ij} -\langle \nabla f,\eta\rangle\_g\delta\_{ij})\\
\end{align}
where we use the transformation of the Levi-Civita connection, i.e.,
$$\overline\nabla\_X Y = \nabla\_XY + (Xf)Y + (Yf)X - \langle X,Y\rangle\_g\nabla f.$$
Thus
\begin{align}
\overline H 
& = \sum\_{i=1}^{n-1}\overline h\_{ii}\\
& = \sum\_{i=1}^{n-1}e^{-f}(h\_{ii} - \langle \nabla f,\eta\rangle\_g)\\
& = e^{-f} (H-(n-1) \langle \nabla f,\eta\rangle\_g),
\end{align}
where $n$ is the dimension of $N,$ so in your case $n-1=2.$


