Let $f(x,y)=\frac{x}{\sqrt{x^2+y^2}}$. Does $f \in W^{1,p}(B)$ for some $p \ge 1$, where $B$ is the open unit disk in $\mathbb{R}^2$?
(I guess we can replace $B$ with a disk with arbitrarily small radius; the singularity is centered at the origin).
Here is what I know:
$f \le 1$ is bounded, so it is in $L^p(B)$ for any $p \ge 1$. Let us consider its derivatives:
$f_x=\frac{1}{\sqrt{x^2+y^2}}(\frac{y^2}{x^2+y^2})\le \frac{1}{\sqrt{x^2+y^2}}$, so for sure $f_x \in L^p(B)$ for $p<2$.
(In fact $f_x \in L^p(B) \iff p<2$).
$f_y=-\frac{yx}{x^2+y^2}(\frac{1}{\sqrt{x^2+y^2}})$, so $|f_y|\le \frac{1}{\sqrt{x^2+y^2}}$, hence $f_y \in L^p(B)$ for $p<2$.
So, is it true that $f \in W^{1,p}(B)$ for some $ 1 \le p <2$?
As shown in the answer to this question of mine, since $f$ is singular at a point only its classical derivatives coincide with the distributional ones. Therefore, by definition, $f\in W^{1,p}(B)$ if and only if $f$ and $\nabla f$ belong to $L^p(B)$. The computations in the question show that this is the case for all $p\in[1, 2)$.