So, I was studying green's functions and in general I understood that if I have an operator $\mathscr{O}$ that acts of a function $h_1(\vec{r})$ such that $$\mathscr{O}h_1(\vec{r})=h_2(\vec{r})$$ Then all I need to do is to find the function, $g(\vec{r})$, on which the operator acts to yield the delta function.
Then I can write, $$h_1(\vec{r})=\int h_2(\vec{\tau})g(\vec{\tau}-\vec{r})\mathrm{d}^3\vec{\tau}$$
Reason being $$\mathscr{O}h_1(\vec{r})=\mathscr{O}\int h_2(\vec{\tau})g(\vec{\tau}-\vec{r})\mathrm{d}^3\vec{\tau}=\int h_2(\vec{\tau})\delta(\vec{\tau}-\vec{r})\mathrm{d}^3\vec{\tau}$$ So far, so good, but then in an effort to solve the Poisson's equation, writes $$V(\vec{r})=\frac{1}{4\pi}\int \frac{\rho(\vec{\tau})}{\epsilon}\frac{1}{|\vec{\tau}-\vec{r}|}\mathrm{d}^3\vec{\tau} $$ Because (and I'm back-calculating) $$-\nabla ^2\left(\frac{1}{4\pi|\vec{r}|}\right)=\delta(\vec{r}).$$ I am unable to understand this move. Is there some mathematical basis for this or just to equate the preconceived notion of potential of point charges does this equation hold good?