Answer
$${\text{Verified}}$$
Work Step by Step
$$\eqalign{
& u\left( {x,y} \right) = y\left( {3{x^2} - {y^2}} \right) \cr
& u\left( {x,y} \right) = \left( {3{x^2}y - {y^3}} \right) \cr
& {\text{Calculate the second partial derivatives }}\frac{{{\partial ^2}u}}{{\partial {x^2}}}{\text{ and }}\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cr
& \frac{{\partial u}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {3{x^2}y - {y^3}} \right] \cr
& \frac{{\partial u}}{{\partial x}} = 6xy \cr
& \frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{\partial }{{\partial x}}\left[ {6xy} \right] = 6y \cr
& \cr
& and \cr
& \cr
& \frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {3{x^2}y - {y^3}} \right] \cr
& \frac{{\partial u}}{{\partial y}} = 3{x^2} - 3{y^2} \cr
& \frac{{{\partial ^2}u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {3{x^2} - 3{y^2}} \right] = - 6y \cr
& \cr
& {\text{Verify the Laplace's equation }}\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0 \cr
& 6y - 6y = 0 \cr
& 0 = 0 \cr
& {\text{Verified}} \cr} $$
