In Landau-Lifschitz (Volume II):
Stress-energy tensor is evaluated in a reference frame where all $g_{ik,l}=0$. From Einstein equations:
$$T^{ik}=\frac{1}{8\pi\kappa}\left(R^{ik}-\frac{1}{2}g^{ik}R\right)=(g^{il}g^{km}-\frac{1}{2}g^{ik}g^{lm})(\partial_n \Gamma^n_{lm}-\partial_m \Gamma^n_{ln}+\Gamma^n_{pn} \Gamma^p_{lm} - \Gamma^n_{pm} \Gamma^p_{ln})$$
Taking into account that all $\Gamma^{l}_{ik}=0$, $T^{ik}$ is derived:
$$T^{ik}=\frac{\partial}{\partial x^l}\Big\{\frac{1}{16\pi\kappa}\frac{1}{-g}\frac{\partial}{\partial x^m}[(-g)(g^{ik}g^{lm}-g^{il}g^{km})]\Big\}$$
Then $T^{ik}$ is denoted by partial derivative of $h^{ikl}$
$$h^{ikl}=\frac{1}{16\pi\kappa}\frac{\partial}{\partial x^m}[(-g)(g^{ik}g^{lm}-g^{il}g^{km})]$$
So, $$(-g)T^{ik}=\frac{\partial{h^{ikl}}}{\partial x^l}$$
Further, stated that in the transformation to arbitrary reference frames $\frac{\partial{h^{ikl}}}{\partial x^l}-(-g)T^{ik}\neq0$ and $(-g)t^{ik}=\frac{\partial{h^{ikl}}}{\partial x^l}-(-g)T^{ik}$. $t^{ik}=\frac{1}{-g}\frac{\partial{h^{ikl}}}{\partial x^l}-T^{ik}$
Where $t^{ik}$ is stress-energy pseudotensor and I want to derive it explicitly.
I have calculated straightforwardly the expression for $t^{ik}$. After a quite long calculation, there is derived the following expression.
Now I am in stuck on how to proceed with the expression. There definitely must not be second-order derivatives of metric tensor. How to proceed? Any hint is appreciated.