I'm new to Relativity and I'm trying to understand this computation: $$ (\nabla^\mu G)_{\mu\nu}=0 $$ where $G=Ric-\frac{1}{2}\mathcal{R}g $ is the Einstein tensor, and $\nabla^\mu=\nabla_{\partial^\mu} $, where $\partial^\mu=g^{\mu\nu}\partial_\nu $ (don't know if it's standard notation).
I use linearity of $\nabla$ and troubles start on the second term, when I try to apply Leibniz rule. I would write
$$ (\nabla^\mu G)_{\mu\nu} = (\partial^\mu\mathcal{R})g_{\mu\nu} + \mathcal{R}(\nabla^\mu g)_{\mu\nu} $$
but my notes apparently use that $$ \mathcal{R}(\nabla^\mu g)_{\mu\nu} = 0 $$ cause they omit it. Is it so? Can anyone explain why?
PS: I might write bull***t, as I'm really a newbie in the field.
$\nabla^\mu g_{\mu\nu}\neq \partial^\mu g_{\mu\nu}$, but it is true that $\nabla^\mu g_{\mu\nu}=0$ because $\nabla^\mu g_{\mu\nu} = g^{\alpha\mu}\nabla_\alpha g_{\mu\nu}$ and $\nabla_\alpha g_{\mu\nu}=0$ since in general relativity you usually work with the Levi-Civita connection, which is metric-compatible. I.e., $\nabla g=0$.
Edit: after discussion in the comments
People like to use the notation $\nabla^\alpha g_{\mu\nu}$ for what should be written as $(\nabla g)^\alpha{}_{\mu\nu}$. Of course, if $\nabla^\alpha g_{\mu\nu}$ were meant to represent $\nabla^\alpha(g_{\mu\nu})$, then it would indeed be equal to $\partial^\alpha g_{\mu\nu}$, since every $g_{\mu\nu}$ is just a scalar function.