$\displaystyle Let\ w=u+iv\ =(x^3-3xy^2+2xy)+\ i(3x^2y-x^2+y^2-y^3)\ \\[2ex]\displaystyle \ u\ =\ \left(x^3-3xy^2+2xy\right)\ and\ v\ =\ (3x^2y-x^2+y^2-y^3\ )\ \\[2ex] \displaystyle u_x=\ 3x^2-3y^2+2y\ \ \\[2ex] \displaystyle u_y=-6xy+2x \\[2ex] \displaystyle v_x=\ \ 6xy-2x=\ -u_y \\[2ex] \displaystyle v_y=\ 3x^2+2y-3y^2=u_x\ \\[2ex] \displaystyle$

$ We\ observe, \\[2ex] \displaystyle i)\ If\ x\not=0\ and\ y\not=0\ then\ \ u_x,u_{y,}v_y,v_x\\[2ex] are\ \ continuous\ functions\ of\ x\ and\ y. \\[2ex] \displaystyle ii)\ u_x=v_y \\[2ex] \displaystyle iii)\ u_y=\ -v_x\ \ \ \ \ \ \ \\[2ex] \displaystyle \therefore{}\ Cauchy\ Reiman's\ Equations\ are\ satisfied. \\[2ex] \displaystyle Since,\\[2ex] w=(x^3-3xy^2+2xy)+\ i(3x^2y-x^2+y^2-y^3\ )\ \\[2ex] \displaystyle \frac{dw}{dx}=u_x+iv_x=(3x^2-3y^2+2y)+i(6xy-2x)\ \\[5ex] \displaystyle By\ Milne\ Thompsons\ method,\ \\[2ex] \displaystyle put\ x=z,\ y=0 \\[2ex] \displaystyle \therefore{}\frac{dw}{dz}=+3z^2\ +i\left(-2z\right)=3z^2-2iz\ \\[2ex]$

$ \begin{align*} \displaystyle f^{'}\left(z\right)&=\ 3z^2-2iz \\[2ex] \displaystyle \therefore{}f\left(z\right)&=\ \int\left(3z^2-2iz\ \right)dz\\[2ex] &=3\ \frac{z^3}{3}-2i\ \frac{z^2}{2}=\ -z^3-iz^2\ \ \ \ \\[2ex] \displaystyle \therefore{}\ f\left(z\right)&=\ z^3-iz^2=z^2(z-i) \\[2ex] \end{align*}$