I want to prove the following two equalities: $$ \oint_{\sigma}(f\nabla g)\cdot T\ dS=\iint_{\Sigma}(\nabla f\times \nabla g)\cdot N\ dA \\ \oint_{\sigma}(f\nabla g+g\nabla f)\cdot T\ dS=0$$
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For the first one:
From the Stokes theorem we have the following: $$\oint_{\sigma}(f\nabla g)\cdot T\ dS=\iint_{\Sigma}(\nabla \times (f\nabla g))\cdot N\ dA$$ So, we have to show that $$\nabla \times (f\nabla g)=\nabla f\times \nabla g$$
We have that $$\nabla \times (f\nabla g)=\text{curl} (f\nabla g)=\text{curl} \left (f\left (g_x, g_y, g_z\right )\right )$$ How can we continue?