 Solution

Write $F=f^{\prime }(R)<0$, so ${\Omega }^2=-F$, and let $(\nabla F{)}^2=g^{\mu \nu }{\partial }_{\mu }F{\partial }_{\nu }F$. Contracting the supplied transformed Ricci tensor with ${\tilde{g}}^{\mu \nu }={\Omega }^{-2}{g}^{\mu \nu }$ givesThe last term contains one trace contribution from ${\nabla }_{\mu }{\nabla }_{\nu }F$ and four from the term proportional to $g_{\mu \nu }$.

In the trace subtraction, ${\tilde{g}}_{\mu \nu }{\Omega }^{-2}=g_{\mu \nu }$. Consequently the transformed Einstein tensor isThese derivatives refer to the original metric. This is the tensor identity used in the Einstein-frame representation of metric f(R) gravity.