Landau-Lifschitz pseudotensor from Einstein equations (part I) In Landau-Lifschitz (Volume II):
Actually, it is not difficult to bring $T^{ik}$ to this form. To do this we start from the field equation
$$T^{ik}=\frac{1}{8\pi\kappa}\left(R^{ik}-\frac{1}{2}g^{ik}R\right)$$
and for $R^{ik}$ we have
$$R^{ik}=\frac{1}{2}g^{im}g^{kp}g^{ln}\Big\{\frac{\partial^2g_{lp} }{\partial x^m\partial x^n}+\frac{\partial^2g_{mn} }{\partial x^l\partial x^p}-\frac{\partial^2g_{ln} }{\partial x^m\partial x^p}-\frac{\partial^2g_{mp} }{\partial x^l\partial x^n}\Big\}$$
(we recall that at the point under consideration, all the $\Gamma^i_{kl}=0$). After simple transformations, the tensor $T^{ik}$ can be put in the form
$$T^{ik}=\frac{\partial}{\partial x^l}\Big\{\frac{1}{16\pi\kappa}\frac{1}{-g}\frac{\partial}{\partial x^m}[(-g)(g^{ik}g^{lm}-g^{il}g^{km})]\Big\}$$
My question is: what is the hint to obtain the expression for $T^{ik}$? After a straightforward calculation, I have a long-expression different from this one.
$$R^{ik}-\frac{1}{2}g^{ik}R=\frac{1}{2}g^{im}g^{kp}g^{nl}\left(g_{lp},_{mn}+g_{mn},_{lp}-g_{mp},_{ln}-g_{ln},_{mp}\right)-$$
$$
-\frac{1}{2}g^{ik}g^{np}g^{ml}\left(g_{lp},_{mn}+g_{mn},_{lp}-g_{np},_{lm}-g_{lm},_{np}\right)
$$
Using the identities in the comments obtained this:
Using the identities I have this $$\frac{1}{2}\left((g^{ik}g^{mn}-g^{im}g^{nk})_{,mn}+g^{nl}g^{ik}_{,nl}+g^{im}g^{kp}g_{ln}g^{ln}_{,mp}-g^{ik}g^{ml}g_{np}g^{np}_{,lm}-g^{ik}g^{np}g_{lm}g^{lm}_{,np}\right)$$
How to proceed then?
 A: The derivation you refer to is done locally in a set of coordinates in which $g_{ik,l} = 0$, which can be also characterized as related to a set of Riemann normal coordinates by a constant linear transform. This means, that you can make operations such as this one
$$g^{ij} (g_{kl,mn}) = (g^{ij}g_{kl,m})_{,n}$$
Another useful identity is $g^{ij}g_{jk,l} = -g^{ij}_{\;\,,l}g_{jk}$ since $g^{ij}g_{jk} = \delta^i_k\,,\; (\delta^i_k)_{,l} = 0$. One last formulas that you need is $g^{ij}g_{ij,k} = - g^{ij}_{\;\,,k}g_{ij} = g_{,k}/g$, where $g$ is the metric determinant (deriving this formula is a nice little exercise on its own).
It is a little bit laborous to get to the Landau-Lifschitz pseudotensor from the bottom up, but with the formulas above it is really just regrouping the terms in the right way. Start with regrouping the Einstein tensor as a divergence of a 3-index tensor using the first set of identities. Then identify terms that can be written as a $1/g \partial_m$ divergence within this tensor using the $g_{,k}/g$ identity and that should do the trick.
A: You have to rearrange and factor terms in a particular way and then use properties of derivatives. For full derivation with explanations on each step, see the blog
