A couple days ago I asked this same question, but i got no answers. Also I asked to some teachers but they couldn't succeed to give me a clear answer. I am asking here again with the hope that someone who knows the answer see the post.
In the paper https://www.researchgate.net/publication/225499707_Contour_Dynamics_of_Incompressible_3-D_Fluids_in_a_Porous_Medium_with_Different_Densities (page 4) I read that if $$ v (x_1,x_2,x_3,t)=-\frac{\rho_2-\rho_1}{4\pi} PV\int_{\mathbb{R}^2} \frac{(y_1,y_2,\nabla f(x-y,t)\cdot y)}{[|y|^2 + (x_3 - f(x-y,t)^2)]^{3/2}}\ dy$$ and for $\varepsilon $ positive and $x=(x_1,x_2)$, we define $$ v^1(x_1,x_2,f(x,t),t) =\lim_{\varepsilon \rightarrow 0} v(x_1-\varepsilon \partial_{x_1} f(x,t) , x_2 - \varepsilon \partial_{x_2} f (x,t), f(x,t) + \varepsilon , t)$$ Then they say $$ v^1(x_1,x_2,f(x,t),t) = v(x_1,x_2,f(x,t),t) + \frac{\rho_2-\rho_1}{2}\frac{\partial_{x_1} f(x,t)(1,0,\partial_{x_1}f(x,t))}{1+(\partial_{x_1}f(x,t))^2 +(\partial_{x_2}f(x,t))ˆ2 } + \frac{\rho_2-\rho_1}{2}\frac{\partial_{x_2} f(x,t)(0,1,\partial_{x_2}f(x,t))}{1+(\partial_{x_1}f(x,t))^2 +(\partial_{x_2}f(x,t))^2 }$$
I dont understand how they get that expression for $v^1$. Please, any help or idea is welcome.
The interface is $x_3=f(x_1)$, and we are approaching it in the normal direction. The normal direction is $(\partial_{x_1}f,\partial_{x_2}f,-1)$ so moving a small distance in that direction means evaluating $v$ at $(x_1-\epsilon\partial_{x_1}f(\vec x,t),x_2-\epsilon\partial_{x_2}f(\vec x,t),x_3+\epsilon)$.
Here's a sketch for the 2D case: