We have the following operators: \begin{align} J^a(z) = \frac{1}{2}\psi_s^{\dagger}(z)\sigma^a_{s s'}\psi_{s'}(z), \hspace{10 mm} \bar{J}^a(z) = \frac{1}{2}\psi_s^{\dagger}(\bar{z})\sigma^a_{s s'}\psi_{s'}(\bar{z}) \end{align} of the right and left -moving SU(2) Kac-Moody algebras (here$ a = x,y,z$ and $s, s' = \uparrow, \downarrow$, \begin{align} [J^a(x),J^b(x')] = \frac{-i}{2\pi}\delta^{ab}\delta'(x-x') + i\epsilon^{abc}J^c(x)\delta(x-x') \end{align} Using the OPE for the fermionic operators $$<\psi_s^{\dagger}(z)\psi(z')_{s'}> = <\psi_s (z) \psi_s^{\dagger}(z')> = \frac{\delta\delta'}{2\pi(z-z')}$$ derive the OPE for the Kac-Moody currents $J^a(z)$ ($\textit{Hint:}$ use $\sigma^a\sigma^b = \delta^{ab} + i\epsilon^{abc}\sigma^c$) \begin{align} :J^a(z)::J^b(z'): ~=~ \frac{i\epsilon^{abc}J^c(z)}{2\pi(z-z')} +\frac{\delta^{ab}}{8\pi^2(z-z')^2} + \dots \end{align} and the same for the $z \rightarrow \bar{z}$ and $J^a\rightarrow \bar{J}^a$ (left-right symmetry).
I've tried some basic stuff, but it is not clear to me how I should begin this problem.
