We defined the Cartan-Killing metric of a Lie-Group $G$ as $$g_{ab}\equiv C_{acd}C_{bdc},$$ where $C_{abc}$ are the structure constants of the Lie-algebra $\mathfrak{g}$. According to my professor it is possible to show that $$\operatorname{tr}(A_aA_b)=g_{ab},$$ where $A_i$ denote the generators in the adjoint representation.
I'm honestly having a really hard time on trying to show that. As far as I understand we have $$A_a=\left.\frac{\partial \tau_A(g)}{\partial \alpha_a}\right|_{g=e},$$ where $\tau_A$ is the adjoint representation and $e$ the unit-element of $G$. I know that one can write $\|A_a\|_{bc}=C_{abc}$. This would then imply that $$g_{ab}=C_{acd}C_{bdc}= \sum_{c,d=1}^3\|A_a\|_{cd}\|A_{b}\|_{dc}.$$ The problem is that I don't really see what this has to do with the trace of the two matrices $A_a$ and $A_b$.
I will write a solution using the notation for structural constants that I prefer:
If $e_1,...,e_N$ is a basis of the Lie algebra ${\mathfrak g}$, then $$ [e_i,e_j]=\sum_{l} c^l_{ij} e_l. $$ Now, $ad(e_i)\circ ad(e_j)(x)= [e_i, [e_j, x]]$ for $x\in {\mathfrak g}$. If $x=e_k$ then $$ [e_i, [e_j, e_k]] = [e_i, \sum_{l} c^l_{jk} e_l]= \sum_{l,m} c^l_{jk} c^m_{il} e_m. $$ Thus, the linear map $ad(e_i)\circ ad(e_j)$ sends $$ e_k\mapsto \sum_{l,m} c^l_{jk} c^m_{il} e_m $$ Computing the trace of this map means setting $k=m$ and taking the sum over $k=1,...N$: $$ tr(ad(e_i)\circ ad(e_j))= \sum_{l,k} c^l_{jk} c^k_{il}. $$ That's your professor's formula.