Finding equation of motion for given Lagrangian with respect to metric Given the following action in d dimensional (0,1,...,d-1) curved spacetime:
$$ S= \int d^dx\sqrt{-g}\mathscr{L}[\chi,\Phi,g^{\mu\nu}] $$
Where:
$$\mathscr{L}=e^{-2\Phi} \left(-\frac{1}{2\kappa^2}[R-2\Lambda+4g^{\mu\nu}\partial_{\mu}\Phi\partial_{\nu}\Phi]+g^{\mu \nu}\partial_{\mu}\chi\partial_{\nu}\chi+\frac{\beta^2g_{\tau\tau}-\beta_H^2}{(2\pi)^2}\chi^2 \right)$$
Where $g^{\mu\nu}$ is the metric $R$ is the ricci scalar relative to this metric, $g_{00}=g_{\tau\tau}=(g^{\tau \tau})^{-1}$,  $\Phi,\chi$ are scalar fields and $\kappa,\beta,\beta_{H},\Lambda$ are constants.
By doing variation by the field $\Phi$ one can obtain the following equation of motion:
$$\frac{\partial \mathscr{L}}{\partial \Phi}-\Delta_{\mu}\left(\frac{\partial \mathscr{L}}{\partial\left(\partial_{\mu}\Phi\right)}\right)=0$$
Where $\Delta_\mu$ is the covariant derivative, more explicitly for our lagrangian this equation reads:
$$R-2\Lambda-4g^{\mu\nu}\partial_{\mu}\Phi\partial_{\nu}\Phi +4g^{\mu\nu}\Delta_{\mu}\Delta_{\nu}\Phi=2\kappa^2\left(g^{\mu \nu}\partial_{\mu}\chi\partial_{\nu}\chi+\frac{\beta^2g_{\tau\tau}-\beta_H^2}{(2\pi)^2}\chi^2\right)$$
Now I want to find another equation of motion, the one with respect to the metric:
$$\frac{\partial \mathscr{L}}{\partial g^{\mu \nu}}-\frac{1}{2}g_{\mu \nu}\mathscr{L} = 0$$
On the one hand:
$$\frac{\partial \mathscr{L}}{\partial g^{\mu \nu}} = e^{-2\Phi} \left(-\frac{1}{2\kappa^2}[R_{\mu\nu}+4\partial_{\mu}\Phi\partial_{\nu}\Phi]+\partial_{\mu}\chi\partial_{\nu}\chi-\frac{\beta^2g^2_{\tau\tau}}{(2\pi)^2} \delta_{\mu}^{\tau}{\delta}_{\nu}^{\tau}\chi^2 \right) $$
Where I use the following identities:
$$\frac{\partial g_{\tau\tau}}{\partial g^{\mu\nu}}=-g^2_{\tau\tau}\delta_{\mu}^{\tau}{\delta}_{\nu}^{\tau}$$
$$R=g^{\mu\nu}R_{\mu\nu}$$
Where $R_{\mu\nu}$ is Ricci tensor.
On the other hand:
$$-\frac{1}{2}g_{\mu \nu}\mathscr{L}=e^{-2\Phi} \left(-\frac{1}{2\kappa^2}\left[-\frac{1}{2}(R-2\Lambda)g_{\mu\nu}-2d\partial_{\mu}\Phi\partial_{\nu}\Phi \right]-\frac{d}{2}\partial_{\mu}\chi\partial_{\nu}\chi-\frac{1}{2}\frac{\beta^2g_{\tau\tau}-\beta_H^2}{(2\pi)^2}g_{\mu \nu}\chi^2 \right) $$
Where this time I used the identity:
$$g^{\mu\nu}g_{\mu \nu}=\delta_{\mu}^{\mu}=d$$
Using the equation of motion of $\Phi$ above, we substitute it in $R-2\Lambda$ of our expression which yields eventually:
$$-\frac{1}{2}g_{\mu \nu}\mathscr{L}= e^{-2\Phi}\left(-\frac{1}{2\kappa^2}[2d\Delta_{\mu}\Delta_{\nu}\Phi]\right) $$
Putting all together
$$\frac{\partial \mathscr{L}}{\partial g^{\mu \nu}}-\frac{1}{2}g_{\mu \nu}\mathscr{L} = 0 \iff R_{\mu\nu}+4\partial_{\mu}\Phi\partial_{\nu}\Phi+2d\Delta_{\mu}\Delta_{\nu}\Phi=2\kappa^2 \left(\partial_\mu\chi\partial_\nu\chi-\frac{\beta^2g^2_{\tau\tau}}{(2\pi)^2} \delta_{\mu}^{\tau}{\delta}_{\nu}^{\tau}\chi^2\right)$$
But according to some papers, the answer should be:
$$R_{\mu\nu}+2\Delta_{\mu}\Delta_{\nu}\Phi=2\kappa^2 \left(\partial_\mu\chi\partial_\nu\chi-\frac{\beta^2g^2_{\tau\tau}}{(2\pi)^2} \delta_{\mu}^{\tau}{\delta}_{\nu}^{\tau}\chi^2\right)$$
Where did I go wrong???
 A: I find where I was mistaken:
In the action that I presented, I assumed:
$$\frac{\partial R}{\partial g^{\mu\nu}}=R_{\mu\nu}$$
This is not the case in my specific action, but for Hilbert-Einstein action is really the case.
instead, the variation for the element which corresponds to $R$ is:
$$S_R=\int d^dx e^{-2\Phi}R$$
$$\delta S_R=\int d^dx e^{-2\Phi}[\delta g^{\mu\nu}R_{\mu\nu}+g^{\mu\nu}\delta R_{\mu\nu}]$$
Where $g^{\mu\nu}\delta R_{\mu\nu}$ expressed by $\delta g^{\mu\nu}$ in the following way
$$g^{\mu\nu}\delta R_{\mu\nu}=-\Delta_{\mu}\Delta_{\nu}(\delta g^{\mu\nu})+g^{\mu\nu}\nabla^2\delta (g^{\mu\nu})$$
doing integration by parts twice, and demanding that the expression will vanish on the boundary of the hyper $d-1$ surface, yield eventually:
$$\delta S_R=\int d^dx e^{-2\Phi}[R_{\mu\nu}-4\partial_{\mu}\Phi\partial_{\nu}\Phi+4g^{\alpha\beta}\partial_{\alpha}\Phi\partial_{\beta}\Phi g_{\mu\nu}+2\Delta_{\mu}\Delta_{\nu}\Phi-2\nabla^2\Phi g_{\mu\nu}]\delta g^{\mu\nu}$$
This means that:
$$\frac{\partial R}{\partial g^{\mu\nu}}=R_{\mu\nu}-4\partial_{\mu}\Phi\partial_{\nu}\Phi+4g^{\alpha\beta}\partial_{\alpha}\Phi\partial_{\beta}\Phi g_{\mu\nu}+2\Delta_{\mu}\Delta_{\nu}\Phi-2\nabla^2\Phi g_{\mu\nu}$$
considering this change will eventually give the correct answer!
