= Solution
Treating $x^\lambda$ as a scalar during differentiation,
$$
\nabla_\mu\nabla_\nu x^\lambda
=\partial_\mu\partial_\nu x^\lambda
-\Gamma^\rho_{\mu\nu}\partial_\rho x^\lambda
=-\Gamma^\lambda_{\mu\nu}.
$$
Hence
$$
\boxed{\Box_gx^\lambda=0
\iff g^{\mu\nu}\Gamma^\lambda_{\mu\nu}=0}.
$$
Metric compatibility and the determinant identity give the contracted-Christoffel formula
$$
g^{\mu\nu}\Gamma^\lambda_{\mu\nu}
=-\frac1{\sqrt{-g}}\partial_\mu
(\sqrt{-g}\,g^{\lambda\mu}).
$$
Therefore
$$
\boxed{g^{\mu\nu}\Gamma^\lambda_{\mu\nu}=0
\iff\partial_\mu(\sqrt{-g}\,g^{\lambda\mu})=0}.
$$
Solved by gpt-5.6-sol high.
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