$\displaystyle Given\ u=e^x\cos{y+x^3-3xy^2} \[2ex]

\displaystyle Partially\ Differentiating\ w.r.t.\ \ \ x \[2ex]

\displaystyle u_x=e^x\cos{y+3x^2-3y^2} \[2ex]

\displaystyle Again,\ Partially\ differentiating\ w.r.t.x \[2ex]

\displaystyle u_{xx}=e^x\cos{y+6x\ }…\left(1\right) \[2ex]

\displaystyle

\[2ex]

\displaystyle Similarly,\ differentiating\ ‘u’\ partially\ w.r.t.\ ‘y’ \[2ex]

\displaystyle u_y=-e^x\sin{y-6xy} \[2ex]

\displaystyle Again,\ differentiating\ partially\ w.r.t.\ ‘y’ \[2ex]

\displaystyle u_{yy}=\ -e^x\cos{y-6x} \[2ex]

\begin{equation} u_{yy}=\ -u_{xx}\ \ \ \ \ \ \ \ \ \ \ \ \ ...\ from\ (1)\[2ex]\therefore{}u_{xx}+u_{yy}=0,\ \ which\ is\ a\ laplace’s\ equation %eq1 \end{equation}\

$ $\displaystyle Since,\ ‘u’\ satisfies\ laplace’s\ equations,\[3ex] hence\ it\ is\ a\ harmonic\ function. \[2ex] $