$\text{At rest :}\vec{R}+\vec{P}+\vec{f}=\vec{0}=3\vec{P}+\vec{f}\text{ hence }\color{red}{k(R-l_0)=3mg}.\\
\text{Mechanic energy of the bead : }E_m=E_0+mv^2/2+k(l-l_0)^2/2+mgz,l=R\cos(\theta/2),z=R(1-\cos\theta)\text{ thus }-v^2=\frac{k}m(R\cos(\theta/2)-l_0)^2+2gR(1-\cos\theta)+2E_0/m\\
\frac{d(-v^2)}{d\theta}=2gR\sin\theta+\frac{k}m(Rl_0\sin'\theta/2)-R^2\sin(\theta/2)=R\sin(\theta/2)[4g\cos(\theta/2)+\frac{k}m(l_0-R)]\\
\text{For }\theta=\frac{2\pi}3,\frac{d(-v^2)}{d\theta}=0\text{ thus }\color{red}{2mg=k(R-l_0)}$