I have been doing some calculation on variation of Ricci's tensor with respect to the metric, that, according with S. Carroll (An Introduction to General Relativity: Spacetime and Geometry, equation 4.62) is \begin{align}\delta R_{\mu \nu}= \nabla_{\rho}(\delta \Gamma_{\mu \nu}^{\rho}) - \nabla_{\nu}(\delta \Gamma_{\lambda \mu}^{\lambda}) \tag{4.62} \end{align} and I got a explicit equation in terms of the metric, which is \begin{equation} g^{\mu \nu} \delta R_{\mu \nu} = -\nabla_{\mu}\nabla_{\nu}\delta g^{\mu\nu} + g_{\mu \nu}\square\delta g^{\mu \nu} \tag{1}\end{equation} that agrees with the result obtained by Prahar here Variation of modified Einstein Hilbert Action. The thing is that, as far as I am concerned, \begin{align} g^{\mu \nu}\delta R_{\mu \nu}= g^{\mu \nu}\frac{\delta R_{\mu \nu}}{\delta g^{\alpha \beta}}\tag{2} \end{align} because it is the variation respect to the metric. So, if we write explicitly the variation $\delta$ as $\frac{\delta}{\delta g^{\alpha \beta}}$ in (1) we would get\begin{align}g^{\mu \nu} \frac{\delta}{\delta g^{\alpha \beta}} R_{\mu \nu} = -\nabla_{\mu}\nabla_{\nu}\frac{\delta}{\delta g^{\alpha \beta}} g^{\mu\nu} + g_{\mu \nu}\square\frac{\delta}{\delta g^{\alpha \beta}} g^{\mu \nu}\end{align} But \begin{align} \frac{\delta g^{\mu \nu}}{\delta g^{\alpha \beta}} = \frac{1}{2} \left(\delta_{\alpha}^{\mu}\delta_{\beta}^{\nu} + \delta_{\alpha}^{\nu}\delta_{\beta}^{\mu} \right) \tag{3} \end{align} and \begin{align}\nabla_{\mu}\delta_{\beta}^{\alpha}=0 \tag{4} \end{align} Therefore, plugging (3) and (4) into (1), we get \begin{align} \frac{\delta R_{\mu \nu}}{\delta g^{\alpha \beta}} = 0. \end{align} Is it true? if not, why does this fall? I think this time notation is messing around with me.
Thanks.