$f(z) = u + iv .................(i)\ Multiplying\ equation\ (i)\ by\ i\ \therefore\ if(z)=iu-v .................(ii)\ Adding\ equation\ (i)\ and\ (ii)\ \ \therefore\ f(z)+if(z)=u+iv+iu-v\ \therefore\ (1+i)f(z)=(u-v)+i(u+v)\ Let\ u-v = P\ and\ u+v=Q\ \therefore\ (1+i)f(z)=P+iQ ...............(iii)\ Since, \ f(z)\ is\ analytic,\ (1+i)\ f(x)\ is\ also\ analytic\ where\ (1+i)\ is\ constant.\ As\ the\ function\ is\ analytic\, CR\ equations\ are\ satisfied\ \therefore\ P_{x} = Q_{y} ................(A)\ \therefore\ P_{y} = -Q_{x} ................(B)\ Differentiating\ equation\ (iii)\ partially\ w.r.t.\ x\ \therefore\ (1+i)f'(z)=P_{x}+iQ_{x}\ \therefore\ (1+i)f'(z)=Q_{y}+iQ_{x} .................(From B)\ Using\ Milne\ Thomson\ method,\ replacing\ x\ with\ z\ and\ y\ with\ 0\ \therefore\ (1+i)f'(z)=[P_{x}]_{(z,0)}-i[P_{y}]_{(z,0)} .................(C) $ Now $P=u-v=\dfrac{x-y}{x^2+4xy-y^2}\ \therefore P_x=\dfrac{(x^2+4xy-y^2)(1)-(x-y)(2x+4y)}{(x^2+4xy-y^2)^2}\ \therefore [P_x]_{(z,0)}=\dfrac{(z^2)-(z)(2z)}{(z^2)^2}\ \therefore [P_x]_{(z,0)}=\dfrac{-z^2}{(z^2)^2}\ \therefore [P_x]_{(z,0)}=\dfrac{-1}{z^2}$ Now $P=u-v=\dfrac{x-y}{x^2+4xy-y^2}\ \therefore P_y=\dfrac{(x^2+4xy-y^2)(-1)-(x-y)(4x-2y)}{(x^2+4xy-y^2)^2}\ \therefore [P_y]_{(z,0)}=\dfrac{(-z^2)-(z)(4z)}{(z^2)^2}\ \therefore [P_y]_{(z,0)}=\dfrac{-5z^2}{(z^2)^2}\ \therefore [P_y]_{(z,0)}=\dfrac{-5}{z^2}$ Substuting $P_x\ and\ P_y$ in equation (C) $\therefore\ (1+i)f'(z)=\Big(\dfrac{-1}{z^2}\Big)-i\Big(\dfrac{-5}{z^2}\Big)\ \therefore f'(z)=\dfrac{(-1+5i)}{(1+i)}.\dfrac{1}{z^2}\ \ Integrating\ both\ sides\ \therefore f(z)=\dfrac{(-1+5i)}{(1+i)}\int\dfrac{1}{z^2}.dz\ \therefore f(z)=\dfrac{(-1+5i)}{(1+i)}.\dfrac{-1}{z}+C $ $\therefore f(z)=\dfrac{(1-5i)}{z(1+i)}$