Thanks to very helpful and detailed answers for my previous question, I understand the functional derivative of the energy with respect to the density. In addition, this functional derivative is equal to zero as follows (this screenshot is a page of the paper: "A bird's-eye view of density-functional theory [PDF]").
Here, this equation means that the sum of the potential terms (external $v_\text{ext}$, Hartree $v_\text{H}$, and exchange-correlation $v_\text{xc}$ (e.g., LDA)) and the functional derivative of the kinetic energy with respect to the density, $δT_s[n]/δn(r)$, is zero. In other words, does this mean that we have zero at all positions as shown in the following figure? Is it right?
Here, the $δT_s[n]/δn(r)$ term remains in the equation. Is it possible to write this term with a clear expression like these potential terms? More precisely, the kinetic energy is a functional of $\psi$ in the Kohn-Sham DFT (i.e., $T_s[\{\psi_i(r)\}_{i=1}^N]$, where $\psi$ is the Kohn-Sham orbital), but can't we write down this derivative using another equation without $δ$?