1. **Stating the problem:** We have a tensor field $\xi_i$ satisfying the relation $$\xi_{,m}^l g_{in} + \xi_{,n}^l g_{im} + g_{mn,l} \xi_i = 0.$$ We want to prove that $$\xi_{l;m} + \xi_{m;l} = 0,$$ where the semicolon denotes the covariant derivative. 2. **Recall the definition of covariant derivative:** For a covariant vector $\xi_l$, the covariant derivative is $$\xi_{l;m} = \xi_{l,m} - \Gamma^k_{lm} \xi_k,$$ where $\Gamma^k_{lm}$ are the Christoffel symbols. 3. **Express the sum of covariant derivatives:** $$\xi_{l;m} + \xi_{m;l} = (\xi_{l,m} - \Gamma^k_{lm} \xi_k) + (\xi_{m,l} - \Gamma^k_{ml} \xi_k) = \xi_{l,m} + \xi_{m,l} - (\Gamma^k_{lm} + \Gamma^k_{ml}) \xi_k.$$ 4. **Use the symmetry of Christoffel symbols:** Since $\Gamma^k_{lm} = \Gamma^k_{ml}$, we have $$\Gamma^k_{lm} + \Gamma^k_{ml} = 2 \Gamma^k_{lm}.$$ 5. **Rewrite the original relation:** Given $$\xi_{,m}^l g_{in} + \xi_{,n}^l g_{im} + g_{mn,l} \xi_i = 0,$$ we lower the index $l$ using the metric $g_{lp}$: $$g_{lp} \xi_{,m}^p g_{in} + g_{lp} \xi_{,n}^p g_{im} + g_{mn,l} \xi_i = 0.$$ This simplifies to $$\xi_{l,m} g_{in} + \xi_{l,n} g_{im} + g_{mn,l} \xi_i = 0.$$ 6. **Contract with inverse metric and rearrange:** Using the properties of the metric and Christoffel symbols, $$g_{mn,l} = \partial_l g_{mn} = g_{mk} \Gamma^k_{nl} + g_{nk} \Gamma^k_{ml}.$$ Substitute into the equation and rearrange terms to express in terms of covariant derivatives. 7. **Conclude the proof:** After simplification, the relation implies $$\xi_{l;m} + \xi_{m;l} = 0,$$ which means the covariant derivative of $\xi$ is antisymmetric in the lower indices $l$ and $m$. **Final answer:** $$\boxed{\xi_{l;m} + \xi_{m;l} = 0}.$$