Variation of the metric w.r.t. the metric in derivation of stress tensor Consider massless free scalar theory
$$S = \int d^4x \sqrt{-g}L = \int d^4x \sqrt{-g} \;g^{cd}\nabla_c\phi \nabla_d \phi.\tag{1}$$
To compute the Hilbert stress-energy tensor we require $\sim\delta S/\delta g_{ab}$. I want to focus on $\delta L/ \delta g_{ab}$.
$$\frac{\delta L}{\delta g_{ab}} = \frac{\delta g^{cd}}{\delta g_{ab}} \nabla_c\phi \nabla_d \phi = - g^{ac} g^{db}\nabla_c\phi \nabla_d \phi= -\nabla^a\phi \nabla^b \phi,\tag{2}$$
where I have used $\delta (g_{ab}g^{bc})=0$ to derive the implied expression for $\delta g^{ab}$ in terms of $\delta g_{ab}$. This is correct according to H. Reall's notes: (eq. 10.31) http://www.damtp.cam.ac.uk/user/hsr1000/part3_gr_lectures_2017.pdf .
However $$L=g^{cd}\nabla_c\phi \nabla_d \phi=g_{cd}\nabla^c\phi \nabla^d \phi\tag{3}$$ so I could also have 
$$\frac{\delta L}{\delta g_{ab}} = \frac{\delta g_{cd}}{\delta g_{ab}} \nabla^c\phi \nabla^d \phi.\tag{4}$$
For the two results to match, it would seem that we need
$$\frac{\delta g_{cd}}{\delta g_{ab}} \sim - \frac{1}{2}(\delta^a_c \delta^b_d +\delta^a_d \delta^b_c  )\;?\tag{5}$$
but I'm pretty sure there should be no minus sign!
 A: Let me show you a similar mistake. 
For simplicity, let us consider Maxwell theory because it is manifest where the metric hides, 
\begin{equation}
S = \frac{1}{4}\int d^4x \sqrt{-g} F_{\rho\sigma}F^{\rho\sigma},  
\end{equation}
where $F_{\mu\nu} := \partial_\mu A_\nu - \partial_\nu A_\mu$. Notice that the definition of $F_{\mu\nu}$ involves no metric. OK, now we try to derive stress-energy tensor by varying the action with respect to the metric. Naively, one may write 
\begin{equation}
\frac{\delta S}{\delta g_{\mu\nu}} 
\ni \frac{1}{4}\int d^4x \sqrt{-g} \frac{\delta (g_{\alpha\rho}g_{\beta\sigma})}{\delta g_{\mu\nu}} F^{\alpha\beta}F^{\rho\sigma}, 
\end{equation}
where I just showed one term in the variation. On the other hand, one may write 
\begin{equation}
\frac{\delta S}{\delta g_{\mu\nu}} 
\ni \frac{1}{4}\int d^4x \sqrt{-g} \frac{\delta (g^{\alpha\rho}g^{\beta\sigma})}{\delta g_{\mu\nu}} F_{\alpha\beta}F_{\rho\sigma}. 
\end{equation}
Then one finds exactly your puzzle if you work out the variations above -- the sign difference. The sign of the last equation above is correct actually. (To check this, you can compute the trace of stress-energy tensor. Maxwell theory in 4d should have a traceless stress-energy tensor). 
The tricky part is, when you write 
\begin{equation}
F_{\rho\sigma}F^{\rho\sigma} = g_{\alpha\rho}g_{\beta\sigma}F^{\alpha\beta}F^{\rho\sigma}, 
\end{equation}
you are hiding some factors of metric into the $F^{\alpha\beta}F^{\rho\sigma}$ part. Recall that $F_{\mu\nu}$ is defined without metric while $F^{\mu\nu}$ is, because it equals $F^{\mu\nu} = g^{\mu\alpha}g^{\nu\beta}F_{\alpha\beta}$. 
A: Yes, there is no minus sign in eq. (5), cf. e.g. this Phys.SE post. 
The error in OP's differentiation (4) wrt. the metric is instead the missing contributions coming from the fact that $\nabla^c \phi=g^{cd}\partial_d\phi$ implicitly depends on the metric.
