Thanks to [this][1] 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}


  [1]: https://math.stackexchange.com/a/20712/336994