Let $0< \alpha< n$, $1 < p < q < \infty$ and $\frac{1}{q}=\frac{1}{p}\frac{\alpha}{n}$. Then: $ \left \ \int_{\mathbb{R}^n} \frac{f(y)dy}{xy^{n\alpha} } \right\_{L^q(\mathbb{R}^n)}\leq$ $C\left\ f\right\ _{L^p(\mathbb{R^n})}$.

This is the standard HardyLittlewoodSobolev inequality(or the theorem of fractional integration).A more direct approach is write $$ \int{f(xy)y^{\alphan}dy}=\int_{y<R}+\int_{y\ge R} $$ For the second term on the RHS,using Holder inequality,and easy to see that it's dominated by $\f\_{L^p}R^{\frac{q}{n}}$. For the first term,one can use the majorizationgiven by the maximal function M,and to see that $$ f\ast y^{\alphan}(x)\leq C(M(f)(x)\cdot R^{\alpha}+\f\_{L^p}\cdot R^{\frac{q}{n}}) $$ Choosing a proper constant R to make the two terms above be equal,and then the desired inequality hold by intergration(note that the maximal operator is bounded on $L^p$ for $1<p<\infty$). 


The function $\vert x\vert^{\alphan}$ is radial homogeneous of degree $\alphan$, so its Fourier transform is radial homogeneous of degree $(\alphan)n=\alpha$ (both locally integrable since $\alpha >0$ and $\alpha>n$ so both are distributions which are easily seen as temperate: Fourier transforms make sense), so your convolution operator is in fact the Fourier multiplier $\vert D_x\vert^{\alpha}$. The question at hand is thus (with homogeneous spaces) $$ \Vert u\Vert_{W^{\alpha,q}}\lesssim \Vert u\Vert_{W^{0,p}},\quad \text{i.e. }W^{0,p}\subset W^{\alpha,q}, $$ which is a particular case of Sobolev injection since $$0>\alpha,\quad p < q,\quad \frac{1}{p}\frac{1}{q}=\frac{\alpha}{n}. $$ 

