Tursinbay said:
Here is the last line$$\frac 1 2 \left( \frac {\partial g_{nm}} {\partial x^l} + \frac {\partial g_{nl}} {\partial x^m} - \frac {\partial g_{lm}} {\partial x^n} \right) \xi^l \left(g^{mk}g^{in}+g^{mi}g^{kn} \right)$$
OK, so let's concentrate on just this subexpression:
$$\left( \frac {\partial g_{nm}} {\partial x^l} + \frac {\partial g_{nl}} {\partial x^m} - \frac {\partial g_{lm}} {\partial x^n} \right) g^{mk}g^{in}~~~~~~~~ (1) $$ The ##m## and ##n## indices are dummy summation indices, so we may interchange them throughout, i.e., $$\left( \frac {\partial g_{mn}} {\partial x^l} + \frac {\partial g_{ml}} {\partial x^n} - \frac {\partial g_{ln}} {\partial x^m} \right) g^{nk}g^{im} ~~~~~~~~ (2)$$ Then, since ##g## is a symmetric tensor, this can be rewritten as $$\left( \frac {\partial g_{nm}} {\partial x^l} + \frac {\partial g_{lm}} {\partial x^n} - \frac {\partial g_{nl}} {\partial x^m} \right) g^{mi}g^{kn}~~~~~~~~ (3)$$ Notice how, if you swap ##k,i## in (3), then add it to (1), the 2nd and 3rd terms in the parentheses cancel? That's the step you were missing in the last line of your post #5.
Actually, there are some shortcut techniques that would help you perform such computations faster.
a) If you have an expression like ##A_{ij} S^{ij}##, where ##A## is antisymmetric and ##S## is symmetric, then the expression is 0.
b) Instead of writing out each expression tediously as you did in post #5, you could have just done it for ##\xi^{k;i}## and then swapped the ##k,i## indices at the end to get the 2nd term.
c) Because the 2nd parenthesis factor is symmetric in ##k,i## (after swapping the ##m,n## dummy indices in the 2nd term), you can immediately know that any term inside the 1st parentheses which is antisymmetric in ##m,n## will not contribute, because of point (a) above.
I hope that helps.
(Btw, be sure to study Samalkhaiat's post #8 carefully. Sometimes it takes several readings and a bit of pen+paper work to get full value from his posts.)