$ \displaystyle We\ know\ z=x+iy \\[2ex] \displaystyle \therefore{}\ \left\vert{}z\right\vert{}\ =\sqrt{x^2+y^2}\not=0 \\[2ex] \displaystyle \therefore{}\ f(z)=\frac{x-iy}{x^2+y^2} \\[2ex] \displaystyle \therefore{}\ f(z)=\frac{x}{x^2+y^2}-\frac{iy}{x^2+y^2} \\[2ex] $

$\displaystyle Comparing\ real\ and\ imaginary\ parts \\[2ex] \displaystyle u=\ \frac{x}{x^2+y^2}\ and\ \\[2ex] v=\ \dfrac{-y}{x^2+y^2}\ \ \ \ \ \ \ \ \ \ \\[2ex] $

$\displaystyle \therefore{}\ ux=\ \frac{\left(x^2+y^2\right)\bullet{}1-x\bullet{}2x}{{\left(x^2+y^2\right)}^2}\\[2ex] \displaystyle ux \dfrac{y^2-x^2}{{\left(x^2+y^2\right)}^2}\ \\[2ex] \displaystyle uy=x\bullet{}\frac{-1}{{\left(x^2+y^2\right)}^2}\bullet{}2y\\[2ex] \displaystyle uy=\frac{-2xy}{{\left(x^2+y^2\right)}^2}\ \ \ \ \\[2ex] \displaystyle vx\ =-y\bullet{}\frac{-1}{{\left(x^2+y^2\right)}^2}\bullet{}2x=\frac{2xy}{{\left(x^2+y^2\right)}^2} \\[2ex] \displaystyle vy=\frac{{\left(x^2+y^2\right)}^2\bullet{}-1-\left(-y\right)\bullet{}2y}{{\left(x^2+y^2\right)}^2}\\[2ex]vy=\dfrac{y^2-x^2}{{\left(x^2+y^2\right)}^2}\ \ \ \ \ \ \ \ \ \ \ \ \\[2ex] $

$\displaystyle we\ observe\ \\[2ex] \displaystyle 1)If\ x\not=0\ and\ y\not=0\ then\ ux,\ uy,\ vx,\ vy\ are\ continuous\ function\ of\ x\ \&\ y. \\[2ex] $

$\displaystyle 2)\ ux\ =vy\ and \\[2ex] \displaystyle 3)\ uy\ =-\ vx\ \\[2ex] \displaystyle \therefore{}Cauchy^{'}s\ Reimann^{'}s\ equation\ are\ satisfied. \\[2ex] \displaystyle \therefore{}\ \ f\left(z\right)=\frac{\bar{z}}{{\left\vert{}z\right\vert{}}^2}\ is\ analytic. \\[2ex] \displaystyle \therefore{}\ \ f^{'}\left(z\right)=\ ux\ +ivx\ \\[2ex] \displaystyle f^{'}(z)=\frac{y^2-x^2}{{\left(x^2+y^2\right)}^2}+i\bullet{}\frac{2xy}{{\left(x^2+y^2\right)}^2} \\[2ex] \displaystyle By\ Milne\ Thompson’s\ method,\\[2ex] Put\ x=z\ and\ y=0 \\[2ex] $

$\displaystyle \therefore{}\ \ f^{'}\left(z\right)=\frac{0^2-z^2}{{\left(z^2+0^2\right)}^2}+i\bullet{}0=\frac{-1}{z^2}\ \ \ \ \\[2ex]$