 Solution

Take the basis $(H_1,H_2,E_{12},E_{21},E_{13},E_{31},E_{23},E_{32})$ of ${\mathfrak{sl}}_3$, where $H_1=\mathrm{diag}(1,-1,0)$, $H_2=\mathrm{diag}(0,1,-1)$ and $E_{ij}$ is a matrix unit. For $D=\mathrm{diag}(1,-2,1)$, $[D,H_r]=0$ and $[D,E_{ij}]=(d_i-d_j)E_{ij}$. Thus the Adjoint representation matrix isSquaring and taking the trace computes the Killing form directly:The independent identity $B(X,Y)=6\mathrm{tr}(XY)$ for the special linear Lie algebra ${\mathfrak{sl}}_3$ also gives $6(1+4+1)=36$.