$\displaystyle With\ u=x^3-3xy^2+3x^2+1,\ \\[2ex] $

$\displaystyle we\ have\\[2ex] Ux=3x^2-3y^2+6x\ \ and\\[2ex] Uy=\ -6xy-6y. \\[2ex] $

$\displaystyle Therefore,\\[2ex] Uxx=^x+6\ and\\[2ex] Uy=-6x-6\ and\ so\\[2ex] Uxx+Uyy=0. \\[2ex] $

$\displaystyle Becasue\ U\ satisfies\ Laplace^{'}\ equation,\ there\ \ exists\ a\ conjugate\ function\ v\left(x,y\right)that \\[2ex] \displaystyle satisfies\ the\ CR\ equations:Ux=Vu,\ Vx=-Uy. \\[2ex] $

(\displaystyle To\ find\ V\ we\ integrate\ these\ \[2ex]

\displaystyle Vy=Ux=3x^2-3y^2+6x=>\ \int(3x^2-3y^2+6x\ )dy+A(x) \[2ex]

\displaystyle Vx=-Uy=6xy+6y=>\ \int(6xy+6y\ )dy+B(y) \[2ex]

)

$\displaystyle where\ A\left(x\right)and\ B\left(y\right)are\ arbitrary\ function\ of\ x\ and\ y\ respectively. \\[2ex] $

$\displaystyle The\ solution\ for\ v \ must\ be\ same\ from\ each\ equation;together\ we\ find\ that\ \\[2ex] \displaystyle 3x^2y-y^3+6xy+c\\[2ex] where\\[2ex] A=c\ and\\[2ex] B=c-y^3\\[2ex] with\ c\ as\ an\ arbitrary\ constant. \\[2ex] \displaystyle In\ combination\ f\left(z\right)=u+iv=z^3+3z^2+C \\[2ex]$