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