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Thanks to this answer.
As $L$ is symmetric, \begin{align} \frac{d (E(S, \textbf{y}))}{d y_u} = 0 + y_l^TL^{l,u} + (L^{u,l}y_l)^T + y_u^TL^{u, u} + (L^{u, u}y_u)^T = 0 \\ 2 y_l^T L^{L, u} + 2y_u^T L^{u, u} = 0 \\ y_u = -(L^{u,u})^{-1}L^{u,l}y_l \end{align}