= Solution
Write $F=f'(R)<0$, 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 $\widetilde g^{\mu\nu}=\Omega^{-2}g^{\mu\nu}$ gives
$$
\boxed{\widetilde R=\Omega^{-2}\left[R+\frac{3}{2F^2}(\nabla F)^2-\frac3F\Box F\right].}
$$
The 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, $\widetilde g_{\mu\nu}\Omega^{-2}=g_{\mu\nu}$. Consequently the transformed <Einstein tensor> is
$$
\boxed{\widetilde G_{\mu\nu}=G_{\mu\nu}
+\frac{3}{2F^2}\left[\partial_\mu F\partial_\nu F-\frac12g_{\mu\nu}(\nabla F)^2\right]
-\frac1F\nabla_\mu\nabla_\nu F+\frac1F g_{\mu\nu}\Box F.}
$$
These derivatives refer to the original metric. This is the tensor identity used in the <Einstein-frame representation of metric f(R) gravity>.
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