Let $\chi_R(x)$ be a smooth function such that $\chi_R(x)=1$ if $|x| \le R$, $\chi_R(x) = 0$ if $|x| \ge 2R$, and $0 \le \chi_R(x) \le 1$ for all $x \in \mathbb R^3$.
How can we show that $\chi_R(x) \nabla \frac 1{|x|} \in L^1(\mathbb R^3) \cap L^{3/2}(\mathbb R^3)$?
In particular, I'm stuck at the $\chi_R(x) \nabla \frac 1{|x|} \in L^{3/2}(\mathbb R^3)$ part, I got $$\int_{\mathbb R^n} \left|\chi_R(x) \nabla \frac 1{|x|}\right|^{3/2} dx \le \int_{B(0,2R)} \left|\nabla \frac 1{|x|}\right|^{3/2} dx = \int_{B(0,2R)}\frac 1{|x|^3} dx =\infty$$
so I got $\infty$ but obviously I'm trying to get less than $\infty$. I think my inequality might be too crude.
Edit: This is more precise and seems to me more plausible, but I'm still not sure if this answers my question: $$\int_{\mathbb R^n} \left|\chi_R(x) \nabla \frac 1{|x|}\right|^{3/2} dx \color{red}{<} \int_{B(0,2R)} \left|\nabla \frac 1{|x|}\right|^{3/2} dx = \int_{B(0,2R)}\frac 1{|x|^3} dx =\infty$$ (I figured it was a strict inequality because $\chi_R(x) < 1$ as $|x|$ is close enough to $2R$. Wishful thinking, though: the first integral is still equal to $\infty$.)
Your function is not in $L^{\frac 3 2} (\Bbb R^3)$!
Split $\Bbb R^3$ into three regions: $\Bbb R^3 = B(0,R) \cup \overline {B(0, 2R) \setminus B(0,R)} \cup (\Bbb R^3 \setminus B(0, 2R))$. The first two intersect only along $|x| = R$, which is a negligible set, therefore you may decompose your integral into a sum of three integrals corresponding to the above union.
The middle set is closed and bounded, therefore compact. Since $\left| \chi _R \nabla \frac 1 {|x|} \right| ^{\frac 3 2}$ is continuous on it, it will also be integrable.
On the third set, $\chi _R$ is identically $0$, so this is trivial.
Let us focus, then, on $\int \limits _{B(0,R)} \left| \chi _R \nabla \frac 1 {|x|} \right| ^{\frac 3 2} \ \Bbb d x$. Introducing the shorthand $B = B(0,R)$ this is, in fact, just
$$\int \limits _B \left| \nabla \frac 1 {|x|} \right| ^{\frac 3 2} \ \Bbb d x = \int \limits _B \left| - \frac {(x_1, x_2, x_3)} {|x|^3} \right| ^{\frac 3 2} \ \Bbb d x = \int \limits _B \frac {|x| ^{\frac 3 2}} {|x| ^{\frac 9 2}} \ \Bbb d x = \int \limits _B \frac 1 {|x| ^3} \ \Bbb d x .$$
Switching to polar coordinates
$$\begin{align} x_1 &= r \cos \beta \cos \alpha \\ x_2 &= r \cos \beta \sin \alpha \\ x_3 &= r \sin \beta \\ r &\in (0,R), \quad \alpha \in (0, 2\pi), \quad \beta \in \left( - \frac \pi 2, \frac \pi 2 \right) , \end{align} $$
the integral becomes
$$\int \limits _0 ^{2 \pi} \int \limits _{-\frac \pi 2} ^{\frac \pi 2} \int \limits _0 ^R \frac 1 {r^3} r^2 \cos \beta \ \Bbb d r \ \Bbb d \beta \ \Bbb d \alpha = 4 \pi \int \limits _0 ^R \frac 1 r \ \Bbb d r = \infty ,$$
therefore the integral on the first set is infinite, hence $\chi _R \nabla \frac 1 {|x|} \notin L^{\frac 3 2} (\Bbb R^3)$.
In fact, to see what are the $p \ge 1$ with $\chi _R \nabla \frac 1 {|x|} \in L^p (\Bbb R^3)$, redo all of the above changing $\frac 3 2$ into $p$: the integral on the first set will become
$$\int \limits _B \frac 1 {|x| ^{2p}} \ \Bbb d x = 4 \pi \int \limits _0 ^R \frac 1 {r ^{2p-2}} \ \Bbb d r ,$$
and this is finite whenever $2p-2 < 1$, i.e. $p < \frac 3 2$ (funnily, your problem concerns precisely the limit case $p = \frac 3 2$).