In the coordinate $\{x^i\}$, the Riemann curvature tensor can be written as $$ R=R_{ijkl}\,dx^i\otimes dx^j\otimes dx^k \otimes dx^l $$ and the Ricci curvature can be written as $$\text{Ric}=R_{ij}\,dx^i\otimes dx^j$$ where$R_{ij}=g^{kl} R_{kilj}\,$. Prove that $$R_{ij;m}=g^{kl} R_{ikjl;m}$$ where $$\theta_{i_1i_2\cdots i_s;m}=\left(D_{\partial_m}\theta\right)(\partial_{i_1},\cdots,\partial_{i_s})\,.$$
My attempt
I got a hint to prove $$R_{jk;m} = (g^{il}R_{ijkl})_{;m} = g^{il}{}_{;m}R_{ijkl} + g^{il}R_{ijkl;m} = g^{il}R_{ijkl;m}\tag{1}$$ and $$g^{il}{}_{;m}=0 \,.\tag{2}$$
As for $(2)$, I am not sure what the meaning of $g^{il}{}_{;m}$ is. I thought that $$ g^{il}{}_{;m}=\left(D_{\partial_m}\,\check g\right)(dx^i,dx^l) $$ where $\check g(\alpha,\beta)=g(\alpha^\sharp, \beta^\sharp)$ is the dual metric in which $\sharp$ denotes the musical isomorphism. Am I right?
I can prove that $Dg=0$ and hence $g_{ij;k}=0$. I hope to use this to prove $(2)$, and I tried to take advantage of $g^{il}g_{lj}=\delta^i_j$. And then I guess that $$(g^{il}g_{lj})_{;m}=g^{il}{}_{;m}g_{lj}+g^{il}g_{lj;m}$$ but I don't know whether this is right, and I don't know how to get $(1)$. They are similar and seem like leibniz rule. Could we conclude that? Any help would be highly appreciated!
The middle equality of your Equation (1) is the second Bianchi identity. A proof has been discussed in this question.
You have the right interpretation of $g^{ij}{}_{;m}$, and are correct that $(g^{ij}g_{jk})_{;m}=g^{ij}{}_{;m}g_{jk} + g^{ij}g_{jk;m}$. Since $Dg=0$ and the identity tensor $\delta_k^i$ is clearly parallel, this shows that $g^{ij}{}_{;m}g_{jk}=0$. Multiply both sides by $g^{kl}$ to see that $g^{il}{}_{;m}=0$.
This should resolve each of your questions in Equation (1).