Answer
$$\eqalign{
& {\text{Slope in direction }}x:0 \cr
& {\text{Slope in direction }}y:4 \cr} $$
Work Step by Step
$$\eqalign{
& z = {x^2}\ln \left( {y + 1} \right) \cr
& {\text{Calculate the partial derivatives }}{z_x}{\text{ and }}{z_y} \cr
& {z_x} = \frac{\partial }{{\partial x}}\left[ {{x^2}\ln \left( {y + 1} \right)} \right] \cr
& {z_x} = \ln \left( {y + 1} \right)\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] \cr
& {z_x} = 2x\ln \left( {y + 1} \right) \cr
& {\text{Find the slope in the }}x{\text{ direction}} \cr
& {z_x}\left( {2,0,0} \right) = 2\left( 0 \right)\ln \left( {y + 10} \right) \cr
& {z_x}\left( {2,0,0} \right) = 0 \cr
& and \cr
& {z_y} = \frac{\partial }{{\partial y}}\left[ {{x^2}\ln \left( {y + 1} \right)} \right] \cr
& {z_y} = {x^2}\frac{\partial }{{\partial y}}\left[ {\ln \left( {y + 1} \right)} \right] \cr
& {z_y} = \frac{{{x^2}}}{{y + 1}} \cr
& {\text{Find the slope in the }}y{\text{ direction}} \cr
& {z_y}\left( {2,0,0} \right) = \frac{{{{\left( 2 \right)}^2}}}{{0 + 1}} \cr
& {z_x}\left( {2,0,0} \right) = 4 \cr} $$
