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The Riesz transform of the function $f(x,y)$ of two variables ($d=2$) reads, $$ \mathcal{R}_x(f(x,y))=\frac{1}{2\pi}\iint \frac{(x-u)f(u,v)}{[(x-u)^2+(y-v)^2]^{3/2}}\,dudv.$$ Let me define $F(x,y)=\partial_x\mathcal{R}_xf(x,y)$. The transformed functions $\tilde{f}(x',y')$ and $\tilde{F}(x',y')$ are defined by the coordinate transformation $$x\to -\frac{-x'-y'}{2 a},y\to -\frac{y'-x'}{2 b},$$ with Jacobian $2ab$. You thus end up with $$\tilde{F}(x',y')=\frac{2}{\pi}\iint \left(a^2(u'-v'-x'+y')^2-2b^2 (u'+v'-x'-y')^2\right)$$ $$\qquad{}\times\left(b^2(u'+v'-x'-y')^2+a^2(u'-v'-x'+y')^2\right)^{-5/2}\tilde{f}(u',v')\,du'dv'.$$ Any reason why you would want to go through this transformation?