= Solution
Use the <curvature sign convention>
$$
R^\rho{}_{\sigma\mu\nu}=\partial_\mu\Gamma^\rho{}_{\nu\sigma}-\partial_\nu\Gamma^\rho{}_{\mu\sigma}+\Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma}-\Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma},
$$
which has the derivative ordering printed in the source for $R^i{}_{0j0}$. A <Christoffel symbol> is of order $\mu/r^2$, so its derivative is of order $\mu/r^3$, whereas a product of two connections is of order $\mu^2/r^4$. Their ratio is $\mu/r\ll1$. The metric is stationary, hence the temporal derivative term vanishes. Consequently
$$
R^i{}_{0j0}\simeq\partial_j\Gamma^i{}_{00}=-\frac12\partial_i\partial_jg_{00}.
$$
Here $g_{00}=-1+2\mu/r$, and $\partial_i\partial_j(1/r)=(3x_ix_j-r^2\delta_{ij})/r^5$. The <weak-field tidal curvature in a stationary metric> is
$$
\boxed{R^i{}_{0j0}=\frac{GM}{c^2r^3}\left(\delta_{ij}-3\frac{x^ix^j}{r^2}\right)+O(\mu^2/r^4).}
$$
\b[Printed sign inconsistency.] The source's positive half-Hessian and its final negative-prefactor expression are the opposites of this result. A concrete check at $(r,0,0)$ is $\Gamma^1{}_{00}\simeq\mu/r^2$, giving $\partial_1\Gamma^1{}_{00}=-2\mu/r^3$, while the printed final expression gives $+2\mu/r^3$. Thus both printed equalities cannot hold with its stated derivative ordering. Reversing the definition of the <Riemann curvature tensor> would give the printed final sign, but would also require reversing the curvature term in the <geodesic deviation> equation. The convention used here and in the next part gives the physical radial stretching and transverse compression.
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