Functional derivative of metric To do functional derivative of some actions, we need to know a functional differential of metrics $g_{\mu \nu}(x)$. 
One of the formulae about that is: $$g_{\mu\nu}\delta g^{\mu\nu} = - g^{\mu\nu} \delta g_{\mu \nu},$$but this confused me, because for arbitrary tensors, it is true that$$A_{\mu\nu}B^{\mu\nu}=A^{\mu\nu}B_{\mu\nu}.$$
What is the difference? 
I suppose $\delta g^{\mu\nu}$ is a tensor because actions must be scalars, but i don't have any proof.
 A: Use   $g^{\mu\nu}g_{\nu\lambda}=\delta^\mu_\nu$. Now for any matrix $G$  and its inverse $G^{-1}$ we have 
$$
0= \delta(\mathbb I)= \delta (G^{-1} G)= (\delta G^{-1}) G + G^{-1}\delta G.
$$
Multiply by $G^{-1}$ from the right to get 
$$
\delta G^{-1} = - G^{-1}\delta G G^{-1}.
$$
Thus 
$$
\delta g^{\mu\nu}= - g^{\mu \rho} \delta g_{\rho\sigma} g^{\sigma \nu}
$$
A: The inverse metric tensor $(g^{-1})^{\mu\nu}$ is often written as $g^{\mu\nu}$ when no confusion can arise. Let us not use this shorthand notation here. OP's formula then reads
$$ \delta(g^{-1})^{\mu\nu}~=~-(g^{-1})^{\mu\lambda}\delta g_{\lambda\kappa}(g^{-1})^{\kappa\nu} .$$
We can now raise and lower indices with the metric/inverse metric, e.g.
$$\delta(g^{-1})^{\mu\nu}~=~-(\delta g)^{\mu\nu},$$
without running into inconsistencies.
A: $\delta g^{\mu \nu}$ is definitely a tensor, but you have to be careful when you have variations of a tensor. To be clear, it is $\delta (g^{\mu \nu})$, not $(\delta g)^{\mu \nu}$. Then you must express $\delta (g^{\mu \nu})$ in terms of $\delta (g_{\mu \nu})$ as given in your first equation. You do this by taking contractions with the metric. This is where the difference comes in. Precisely, we have
\begin{aligned}
\delta (g^{\mu \nu}) &= \delta (g^{\alpha \mu} g^{\beta \nu} g_{\alpha \beta})\\
&= \delta (g^{\alpha \mu}) \bar{g}^{\beta \nu} \bar{g}_{\alpha \beta} + \bar{g}^{\alpha \mu} \delta(g^{\beta \nu}) \bar{g}_{\alpha \beta} + \bar{g}^{\alpha \mu} \bar{g}^{\beta \nu} \delta(g_{\alpha \beta})
\end{aligned}
where an overbar on the metric $\bar{g}_{\mu \nu}$ indicates that the metric takes its background value.
