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Set $z=x+y^2$. Then $$ (xy+x^3,y^2,xy^2+x^5)=(xz,(z-x^2)^2,xz(y-x^2)+2x^5) =(xz,x^5,z^2+x^4) $$ and similarly $(xy+x^3,y^2+ux^4,xy^2+x^5)=(xz,x^5,z^2+(1+u)x^4)$. So, after setting $x'=x\root4\of{1+u}$ we get the required isomorphism.