Let $G(|x-y|)=\frac{1}{3\cdot \text{Vol}(\mathbb S^2)|x-y|}$. Then it's written in my course that $$-\Delta_x G(|x-y|)=\delta _{y}.$$
Does it mean that $$\Delta _xG(|x-y|)=0$$ when $x\neq y$ and $0$ when $x=y$ ? But I know that $\delta_y$ is a distribution... but I always have difficulties to understand what it really means those distribution. Because as distribution, it looks that $$-\int_A\Delta _{y}G(|x-y|)dx=\begin{cases}1&y\in A\\ 0&y\notin A\end{cases},$$
but it doesn't really make sense... since we want to solve the equation $\Delta u=0$. I really have a lot of truble to understand what they want (btw, in the famous book PDE's of Evans, they also write this equation).
As you already suspected, the equation $$-\Delta_x G(|x-y|)=\delta_x \tag{1}$$ has to be read in a distributional sense. For this to make sense, note that the function $$ g_x\colon \Bbb R^3 \to \Bbb R, \quad y\mapsto G(|x-y|),$$ is locally integrable. So it defines the regular distribution $$ T_x\colon C_c^\infty(\Bbb R^3) \to \Bbb R, \quad \varphi\mapsto \int_{\Bbb R^3}g_x(y)\varphi(y)dx.$$ Equation (1) is supposed to express that $$-\Delta_x (T_x \varphi)= \delta_x(\varphi)=\varphi(x),$$ i.e. $u(x):=T_x \varphi$ solves the Poisson equation $-\Delta u = \varphi$.