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\(x+y=\cfrac{\sqrt3+1}{\sqrt3-1}+\cfrac{\sqrt3-1}{\sqrt3+1}\)
\(=\cfrac{(\sqrt3+1)^2+(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)}\)
\(=\cfrac{2[(\sqrt3)^2+(1)^2]}{3-1}\)
\(=\cfrac{2[3+1]}{2}\)
\(=\cfrac{8}{2}\)
\(=4\)

\(xy=\cfrac{\sqrt3+1}{\sqrt3-1}\times \cfrac{\sqrt3-1}{\sqrt3+1}=1\)

\(\therefore \cfrac{x^2-xy+y^2}{x^2+xy+y^2}\)
\(=\cfrac{x^2+y^2-xy}{x^2+y^2+xy}\)
\(=\cfrac{(x+y)^2-2xy-xy}{(x+y)^2-2xy+xy}\)
\(=\cfrac{(x+y)^2-3xy}{(x+y)^2-xy}\)
\(=\cfrac{(4)^2-3\times 1}{(4)^2-1}\)
\(=\cfrac{16-3}{16-1}\)
\(=\cfrac{13}{15}\) (Answer)