Initial task is to find out, for which $s$ stands $u=||x||^s \in H^1(\Omega)$, where $\Omega = B(1,0)\subset\mathbb{R}^n$ and $H^1(\Omega)$ is a Sobolev space $W^{1,2}(\Omega)$. As to prove this, we need to prove that every weak partial derivative of the first order and the function itself are in $L^2(\Omega)$. For partial derivatives we get:
$$\partial _i u = s||x||^{s-2} x_i,$$
and everywhere but zero weak partial derivative has to coincide with this partial derivative ($0$ as a zero set doesn't affect anything). So we get:
$$\begin{array} &||\partial_iu||_{L^2(\Omega)}&=&\left(\int_\Omega (s||x||^{s-2}x_i)^2\right) ^{\frac{1}{2}}\\ &=&|s|\left(\int_\Omega(||x||^{s-2}x_i)^2\right)^\frac12 \end{array}$$
How can I tell now whether integral is finite of not?
The function $x\mapsto \|x\|^{2s-4}x_i^2$ is homogeneous of degree $2s-4+2=2s-2$. So it is integrable on the unit ball if and only if $2s-2>-n$. This is a special case of general fact about homogeneous functions. (I.e., those with $f(tx)=t^df(x)$ for all $x$ and all $t>0$.)
Indeed, suppose $f$ is homogeneous of degree $d$ and is not zero a.e. Decompose the unit ball into dyadic shells $A_k=\{x:2^{-k-1}<|x|<2^{-k}\}$ for $k=0,1,2,3,\dots$. Let $I=\int_{A_0}|f|$. By the change of variable $x=2^{-k}y$, $$\int_{A_k}|f(x)|\,dx = \int_{A_0}2^{-dk}|f(y)|2^{-nk}\,dy = 2^{-(d+n)k}I$$ The sum over $k$ converges if and only if $d+n>0$.