We are given that a state, say \(|\psi\rangle\), is an eigenstate of the operators \(L_x\) and \(L_y\), that is;
\[\begin{equation} L_x|\psi\rangle=l_x|\psi\rangle,\,\,\,\, L_y|\psi\rangle=l_y|\psi\rangle \tag{1} \end{equation}\] We also know that; \[\begin{equation} [L_x,L_y]=i \hbar L_z. \tag{2} \end{equation}\] Note that this is an operator relation which is valid on all kets. We can apply this operator equality onto our ket \(|\psi\rangle\). \[\begin{equation} [L_x,L_y]|\psi\rangle=L_x L_y|\psi\rangle-L_y L_x|\psi\rangle=(l_x l_y-l_y l_x)|\psi\rangle=0. \tag{3} \end{equation}\] This implies, using Eq. (2) \[\begin{equation} L_z|\psi\rangle=0. \tag{4} \end{equation}\] One can repeat the same steps; \[\begin{equation} [L_z,L_x]|\psi\rangle=L_z L_x|\psi\rangle-L_x L_z|\psi\rangle=0=i \hbar L_y|\psi\rangle. \tag{5} \end{equation}\] The second method to solve the problem is to use uncertainty relations, which can be read from Shankar Chapter 9[1],
\[\begin{equation} (\Delta \Omega)^2 (\Delta\Lambda)^2\geq \frac{1}{4}(\langle\psi|\{\hat\Omega,\hat\Lambda\}|\psi\rangle)^2+\frac{1}{4}|\langle\psi|[\Lambda,\Omega]|\psi\rangle|^2 \tag{6} \end{equation}\] As the state we are considering is an eigenstate of operator \(L_x\) (\(L_y\)), associated uncertainty \(\Delta L_x\) (\(\Delta L_y\)) is zero. That means if we choose the arbitrary operators \(\Lambda\) and \(\Omega\) as \(L_x\) and \(L_z\), left hand side of Eq. (6) is zero. As both terms on right hand side are nonnegative, they both have to vanish, which implies that \(\langle\psi|L_y|\psi\rangle=0\). It is worth to emphasize that \(\langle\psi|\Theta|\psi\rangle=0\) does not imply \(\Theta|\psi\rangle=0\) for all \(\Theta\) and \(|\psi\rangle\). But in our case \(\langle\psi|L_y|\psi\rangle=0=l_y \langle\psi|\psi\rangle\) which implies \(l_y=0\), since \(\langle\psi|\psi\rangle\neq 0\). To prove that \(l_x=0\), we need to choose the arbitrary operators \(\Lambda\) and \(\Omega\) as \(L_y\) and \(L_z\), and follow the same steps.