I know that
if $f\colon \mathbb{R}^3 \to \mathbb{R}$ is integrable on any measurable, according to the usual $n$-dimensional Lebesgue measure $\mu_y$, and bounded subset of $\mathbb{R}^3$, and if $g \in C^k(\mathbb{R}^3)$ is compactly supported, then the function$$h:x\mapsto \int_{\mathbb{R}^3} f(y-x)g(y)\,d\mu_y$$belongs to $C^k(\mathbb{R}^3)$, and its partial derivatives of order $\leqslant k$ are given by
$$D^{\alpha} h(x) = \int_{\mathbb{R}^3} f(y-x)D^{\alpha} g(y)\,d\mu_y$$
whose proof, whose author I thank again, is found here. An analogous result holds if $g\in C^k(\mathbb{R}^3)$ is bounded and $f\in L^1(\mathbb{R}^3)$, as this other proof, whose author I also thank again, shows.
I would say that, if $$f(z):z\mapsto \frac{1}{\|z\|}$$ and the partial derivatives $D^\alpha g$, with $0\le\sum_{i=1}^n\alpha_i\le k$, of $g$ decay more rapidly than $\|y\|^{-(2+\varepsilon)}$, $\varepsilon>0$, i.e. if $$\exists C>0\,\exists\varepsilon>0:\forall y\in\mathbb{R}^3\quad\|x\|^{2+\varepsilon}|D^\alpha g(y)|< C$$(a condition weaker than belonging to the Schwartz space, but satisfied by Schwartz functions), then, for all $x\in\mathbb{R}^3$, the function defined by $$y\mapsto f(y-x)D^\alpha g(y) d\mu_y$$is Lebesgue summable on $\mathbb{R}^3$. Can we extend the lemma above to this $f$, i.e. is it true and, if it is, how can it be proved (if possible by using elementary real analysis tools like basic theorems about Lebesgue integrations) that $$D_x^\alpha\int_{\mathbb{R}^3} \frac{g(y)}{\|y-x\|} d\mu_y=\int_{\mathbb{R}^3}\frac{D_y^\alpha g(y)}{\|y-x\|} d\mu_y?$$I am not sure whether that holds and how can it be proved because the $\chi_{L-x_0}$ used in the first linked proof cannot be used this time. I heartily thank any answerer.