I'm trying to understand the derivation of a pretty well-known formula (4.15) in https://arxiv.org/pdf/hep-th/9701123.pdf, namely \begin{equation} \frac{\partial^3 \mathcal{F}}{\partial a_I\partial a_J\partial a_K} = Res_{d\omega = 0}\frac{dW_IdW_JdW_K}{d\omega d\lambda}. \end{equation} Where $\mathcal{F}$ is the Seiberg-Witten prepotential, $a^D_I = \partial \mathcal{F}/\partial a_I$. $dW_I$ are normalised basis 1-form on the SW curve, $dW_I = (\partial/\partial a_I)dS_{SW}$. The Seiberg-Witten differential is given by $dS_{SW} = \lambda d\omega, d\omega = dw/w$. The Seiberg-Witten curve is defined by $\mathcal{P}(\lambda,w) = (w + 1/w) - P_N(\lambda) = 0$.
I managed to follow the derivation up to (4.13) \begin{equation} \frac{\partial dv_L}{\partial s_M} = -\Big[\Big(\frac{(\partial \mathcal{P}/\partial s_M)(\partial \mathcal{P}/\partial s_L)}{\mathcal{P}'} - \frac{\partial^2\mathcal{P}}{\partial s_M\partial s_L}\Big)\Big]\frac{d\omega}{\mathcal{P}'} \end{equation} where $\{s_I\}$ is the moduli space coordinates and $dv_I = (\partial/\partial s_I) dS_{SW}$. However I don't understand the next step which is to substitute this result back into (4.11). Assuming I don't care about $\Sigma_{KLM}$ part for now(?), (it is zero if all $b$-periods are closed and I'm happy with that assumption), how do I show that \begin{equation} Res (du_K \frac{\partial v_L}{\partial s_M}) = -Res_{d\omega = 0}\frac{(\partial \mathcal{P}/\partial s_K)(\partial \mathcal{P}/\partial s_L)(\partial \mathcal{P}/\partial s_M)}{(\mathcal{P}')^3}\frac{d\omega^2}{d\lambda}? \end{equation} I think my main problem is I don't know how to integrate $\partial dv_K/\partial s_M$ to get $\partial v_L/\partial s_M$.