Variation of $\log(-\det g_{\mu\nu})$ for Einstein -Hilbert action In Einstein-Hilbert variation is needed the variation of the determinant of a metric tensor. For convenience, firstly evaluated the variation of $\log(-\det g_{\mu\nu})$ considering $\det g_{\mu\nu}<0$
$$\delta\log( -\det g_{\mu\nu})=\log[-\det(g_{\mu\nu}+\delta g_{\mu\nu})]-\log[-\det g_{\mu\nu}]=$$
$$
\log\det g^{\mu\nu}(g_{\mu\nu}+\delta g_{\mu\nu})=\log\det(\hat {1}+g^{\mu\nu}\delta g_{\mu\nu})=
$$
$$
Tr\log(\hat {1}+g^{\mu\nu}\delta g_{\mu\nu})=Tr(g^{\mu\nu}\delta g_{\mu\nu})
$$
How to proceed then? Is $Tr(g^{\mu\nu}\delta g_{\mu\nu})=g^{\mu\nu}\delta g_{\mu\nu}?$ Following that $Tr$ (Scalar)=Scalar the identity must be true. Is that correct?
 A: Use the chain rule to evaluate the variation of the logarithm: let $F[X]$ be a functional of $X = X(\tau)$. Then
$$\delta \log(F[X]) = \frac{1}{F[X]} \delta F[X]$$
In your example (suppressing indices on the metric):
$$ \delta \log(-\det(g)) = \frac{1}{\det(g)} \delta \det(g) $$
You should then use Jacobi's formula:
$$ \delta \det(g) = \det(g)g^{\mu\nu}\delta g_{\mu\nu}$$
edit: I have misunderstood the point of the question: it seems like OP was looking to derive this final result, and their claim in the final line is true and verifies Jacobi's result.
A: Using indices can be confusing. You are actually calling $g_{\mu \nu}$ the metric tensor as a matrix, not a particular element of this matrix. 
Your derivation is correct until the second last step. It becomes clearer if you use the substitution/notation
\begin{align}
g_{\mu \nu} &\to g \\
g^{\mu \nu} &\to g^{-1} \\
\delta g_{\mu \nu} &\to \delta g
\end{align}
At this point your last equation becomes
$$
\mathrm{Tr}{ (g^{-1} \delta g )} = g^{lm} {\delta g_{lm}} \,
$$
where I used summation notation for repeated indexes. Now taking the derivative with respect to the variation of $g$ should be obvious.
A: To account for the minus sign, one can e.g. use the following trick
$$ -\det g~\stackrel{(2)}{=}~+\det M, \tag{1}$$
where the matrix $$M^{\mu}{}_{\nu}~:=~(\eta^{-1})^{\mu\lambda} g_{\lambda\nu}\tag{2} $$
has positive determinant, and $\eta_{\mu\nu}$ is a fixed fiducial reference metric of determinant $-1$, e.g. the  Minkowski metric. Then the infinitesimal variation is
$$\begin{align} \delta\ln(-\det g)~\stackrel{(1)}{=}~& \delta\ln\det M\cr ~=~&\delta {\rm tr}\ln M \cr 
~=~&{\rm tr}( M^{-1}\delta M)\cr 
~\stackrel{(2)}{=}~&{\rm tr}( g^{-1}\eta\eta^{-1} \delta g)\cr
~=~&{\rm tr}(g^{-1} \delta g)\cr
~=~&(g^{-1} \delta g)^{\mu}{}_{\mu}\cr
~=~&( g^{-1})^{\mu\nu} \delta g_{\nu\mu}. \end{align}\tag{3}$$
