OK, I can't figure out why I can't get this right:
$$\nabla_y \left( \frac{1}{|x-y|} \right)= \frac{x-y}{|x-y|^3},$$
right? I've checked the calculation several times, although this student-written article has a minus sign there. (Maybe they're confusing $\nabla_x$.)
Now, supposing I'm right, I'm trying to show that for $u$ a harmonic function in $\Omega \subset \mathbb{R}^3$ a smooth bounded region with boundary $S$, we have
$$\frac1{4\pi}\int_{y\in S}\left( u(y) \frac{\cos \beta}{|x-y|^2}+\frac{1}{|x-y|}\frac{\partial u}{\partial \nu}(y) \right)dS(y), \tag{1}$$
where $\beta$ is the angle between $\nu$ and $x-y$. We use $$\int_{\Omega}u\Delta v - v\Delta udV = \int_S \left( u\frac{\partial v}{\partial \nu}- v\frac{\partial u}{\partial \nu} \right). \tag{2}$$ Let $v = \frac{1}{4\pi}\frac{1}{|x-y|}$. This is the fundamental solution in $\mathbb{R}^3$, so $\Delta v=-\delta_x$. Then the left hand side of (2) is $-u(x)$. The right hand side of (2) should be (1) except with a minus sign on the first term, right?
Either I've made a mistake somewhere, or both the question and the student-written article linked above are wrong.
Yes, $\nabla_y \left( \frac{1}{|x-y|} \right)= \frac{x-y}{|x-y|^3}$ is correct. It says that the gradient points from $y$ to $x$, as it should: the function increases toward $x$.
When unsure about $\pm$, try a simple example. Let $u\equiv 1$ on the unit ball $\Omega$, with $x=0$. Does the formula in the article work? No, because the angle between outward normal $n$ and the vector $x-x'$ is $\pi$, making $\cos\beta=-1$. So there.