Answer
$${z_x} = \frac{{2x}}{{{x^2} + {y^2} + 1}}{\text{ and }}{z_y} = \frac{{2y}}{{{x^2} + {y^2} + 1}}$$
Work Step by Step
$$\eqalign{
& z = \ln \left( {{x^2} + {y^2} + 1} \right) \cr
& {\text{Calculate }}{z_x}{\text{ treating }}y{\text{ as a constant}} \cr
& {z_x} = \frac{\partial }{{\partial x}}\left[ {\ln \left( {{x^2} + {y^2} + 1} \right)} \right] \cr
& {z_x} = \frac{{\frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2} + 1} \right]}}{{{x^2} + {y^2} + 1}} \cr
& {z_x} = \frac{{2x}}{{{x^2} + {y^2} + 1}} \cr
& {\text{Calculate }}{z_y}{\text{ treating }}x{\text{ as a constant}} \cr
& {z_y} = \frac{\partial }{{\partial y}}\left[ {\ln \left( {{x^2} + {y^2} + 1} \right)} \right] \cr
& {z_y} = \frac{{\frac{\partial }{{\partial y}}\left[ {{x^2} + {y^2} + 1} \right]}}{{{x^2} + {y^2} + 1}} \cr
& {z_y} = \frac{{2y}}{{{x^2} + {y^2} + 1}} \cr} $$
