1. Statement of the problem. We need to derive the identity $$V^{-1} \sum_{\star} \left( \frac{\partial \varphi}{\partial k} \times \frac{\partial \psi}{\partial k} \right) = 0$$ 2. Key identity and rules. We use the vector calculus identity for the curl of a scalar times a gradient. $$\nabla_k \times (\varphi \nabla_k \psi) = \nabla_k \varphi \times \nabla_k \psi$$ This identity follows from the product rule for curl and from $\nabla_k \times \nabla_k \psi = 0$. 3. Replace the discrete sum by a volume integral and apply the identity. Assuming $V^{-1} \sum_{\star}$ represents the average over k-space volume, we write $$V^{-1} \sum_{\star} \left( \nabla_k \varphi \times \nabla_k \psi \right) = V^{-1} \int_V \left( \nabla_k \times (\varphi \nabla_k \psi) \right) \, d^3k$$ 4. Apply Stokes' theorem. The volume integral of a curl equals the surface integral of the field on the boundary, $$\int_V (\nabla_k \times A) \, dV = \oint_{\partial V} (\hat n \times A) \, dS$$ With $A=\varphi \nabla_k \psi$ we get $$V^{-1} \int_V \left( \nabla_k \times (\varphi \nabla_k \psi) \right) \, d^3k = V^{-1} \oint_{\partial V} \left( \hat n \times (\varphi \nabla_k \psi) \right) \, dS$$ 5. Use periodic boundary conditions. For a Brillouin zone or any domain with periodic boundary conditions the surface integral cancels pairwise and vanishes. 6. Conclusion. Therefore $$V^{-1} \sum_{\star} \left( \frac{\partial \varphi}{\partial k} \times \frac{\partial \psi}{\partial k} \right) = 0$$ This completes the derivation.