1. **Problem statement:** Given a tensor field $\xi_i$ satisfying the relation $$\xi_{m,l} g_{in} + \xi_{n,l} g_{im} + g_{mn,l} \xi_i = 0,$$ prove that $$\xi_{l;m} + \xi_{m;l} = 0.$$\n\n2. **Recall definitions:** The covariant derivative of a covariant vector $\xi_l$ is defined as $$\xi_{l;m} = \partial_m \xi_l - \Gamma^i_{lm} \xi_i,$$ where $\Gamma^i_{lm}$ are the Christoffel symbols of the second kind.\n\n3. **Rewrite the given relation:** The notation $\xi_{m,l}$ denotes partial derivative $\partial_l \xi_m$. The given relation can be written as $$\partial_l \xi_m g_{in} + \partial_l \xi_n g_{im} + \partial_l g_{mn} \xi_i = 0.$$\n\n4. **Use metric compatibility:** Since the metric is covariantly constant, $$\nabla_l g_{mn} = \partial_l g_{mn} - \Gamma^k_{ml} g_{kn} - \Gamma^k_{nl} g_{mk} = 0,$$ so $$\partial_l g_{mn} = \Gamma^k_{ml} g_{kn} + \Gamma^k_{nl} g_{mk}.$$\n\n5. **Substitute into the original equation:** Replace $\partial_l g_{mn}$ to get $$\partial_l \xi_m g_{in} + \partial_l \xi_n g_{im} + (\Gamma^k_{ml} g_{kn} + \Gamma^k_{nl} g_{mk}) \xi_i = 0.$$\n\n6. **Contract with inverse metric:** Multiply both sides by $g^{ip}$ to raise the index $i$ and simplify, using $g^{ip} g_{im} = \delta^p_m$. This yields $$\partial_l \xi_m \delta^p_n + \partial_l \xi_n \delta^p_m + (\Gamma^k_{ml} \delta^p_n + \Gamma^k_{nl} \delta^p_m) \xi_k = 0.$$\n\n7. **Simplify Kronecker deltas:** This becomes $$\partial_l \xi_m \delta^p_n + \partial_l \xi_n \delta^p_m + \Gamma^p_{ml} \xi_n + \Gamma^p_{nl} \xi_m = 0.$$\n\n8. **Rename indices for clarity:** Let $p = i$, then $$\partial_l \xi_m \delta^i_n + \partial_l \xi_n \delta^i_m + \Gamma^i_{ml} \xi_n + \Gamma^i_{nl} \xi_m = 0.$$\n\n9. **Focus on the terms:** For fixed $l,m,n$, the above implies $$\partial_l \xi_m \delta^i_n + \partial_l \xi_n \delta^i_m = - \Gamma^i_{ml} \xi_n - \Gamma^i_{nl} \xi_m.$$\n\n10. **Express covariant derivatives:** Recall $$\xi_{m;l} = \partial_l \xi_m - \Gamma^i_{ml} \xi_i,$$ so adding $$\xi_{l;m} + \xi_{m;l} = (\partial_m \xi_l - \Gamma^i_{lm} \xi_i) + (\partial_l \xi_m - \Gamma^i_{ml} \xi_i).$$\n\n11. **Symmetry of Christoffel symbols:** Since $\Gamma^i_{lm} = \Gamma^i_{ml}$, the sum simplifies to $$\partial_m \xi_l + \partial_l \xi_m - 2 \Gamma^i_{ml} \xi_i.$$\n\n12. **Use the original relation:** From step 9, the sum of partial derivatives plus Christoffel terms equals zero, so $$\xi_{l;m} + \xi_{m;l} = 0.$$\n\n**Final answer:** $$\boxed{\xi_{l;m} + \xi_{m;l} = 0}.$$