I tried to prove the statement below using the properties of the Lie bracket and the isometry. I could not prove it, the pull back function ($ f _ {*} $) does not appear by multiplying the $ M $ manifolds connection. Can someone help me please.
Let $\left( M, g \right)$ and $\left(M′, g′\right)$ be Riemannian manifolds, denote by $ \nabla $ and $\nabla^{'}$ the corresponding Levi-Civitá connections, and let $f : M \rightarrow M′$ be an isometry. It follows from the Koszul formula that $f$ maps $\nabla$ to $\nabla^{'}$ is the sense that
$$ {\nabla^{'}}_{f_{*}X} f_{*}Y = f_{*}\left( \nabla_{X} Y \right). $$
Koszul formula is given by \begin{align} 2g ( \nabla_{X} Y, Z) = Xg(Y, Z) + Y g(Z, X) − Zg(X, Y ) - g([X, Y ], Z) - g([Z, X], Y ) - g([Z, Y ], X) \end{align}
Thank!