= Solution
For a point mass, $\Phi=-GM/r$, and part (b) gives
$$
h_{00}=\frac{2GM}{r}.
$$
The <linearized Ricci tensor and scalar> obey
$$
R=\partial_\mu\partial_\nu h^{\mu\nu}-\mathop{\Box}h.
$$
For a static perturbation with $h_{0i}=0$, this becomes
$$
R=\partial_i\partial_jh_{ij}+\nabla^2h_{00}-\nabla^2h_{ii}.
$$
Away from the source, $\nabla^2h_{00}=0$. For $h_{ij}=f(r)x_i x_j$, one has $h_{ii}=r^2f$, so the two supplied identities give
$$
R=
\left(r^2f''+8rf'+12f\right)
-\left(r^2f''+6rf'+6f\right)
=2rf'+6f.
$$
The vacuum trace equation $R=0$ therefore reduces to
$$
rf'+3f=0,
$$
whose general solution for $r>0$ is
$$
\boxed{f(r)=\frac{C}{r^3}}.
$$
The scalar equation alone leaves $C$ arbitrary. Imposing the remaining vacuum components $R_{ij}=0$ and matching the mass fixes $C=2GM$, as in the <linearized point-mass metric in radial Cartesian form>.
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