Answer
$$\frac{{\partial z}}{{\partial r}} = \frac{{2r{{\cos }^2}\theta }}{{{r^2}{{\cos }^2}\theta + 1}}{\text{ and }}\frac{{\partial z}}{{\partial \theta }} = \frac{{ - 2{r^2}\cos \theta \sin \theta }}{{{r^2}{{\cos }^2}\theta + 1}}$$
Work Step by Step
$$\eqalign{
& z = \ln \left( {{x^2} + 1} \right);\,\,\,\,\,\,\,\,\,x = r\cos \theta \cr
& {\text{Substitute }}r\cos \theta {\text{ for }}x{\text{ into the function }}z = \ln \left( {{x^2} + 1} \right) \cr
& z = \ln \left( {{{\left( {r\cos \theta } \right)}^2} + 1} \right) \cr
& z = \ln \left( {{r^2}{{\cos }^2}\theta + 1} \right) \cr
& {\text{Find the partial derivatives }}\frac{{\partial z}}{{\partial r}}{\text{ and }}\frac{{\partial z}}{{\partial \theta }} \cr
& \frac{{\partial z}}{{\partial r}} = \frac{\partial }{{\partial r}}\left[ {\ln \left( {{r^2}{{\cos }^2}\theta + 1} \right)} \right] \cr
& {\text{Treat }}\theta {\text{ as a constant}} \cr
& \frac{{\partial z}}{{\partial r}} = \frac{1}{{{r^2}{{\cos }^2}\theta + 1}}\frac{\partial }{{\partial r}}\left[ {{r^2}{{\cos }^2}\theta + 1} \right] \cr
& \frac{{\partial z}}{{\partial r}} = \frac{1}{{{r^2}{{\cos }^2}\theta + 1}}\left( {2r{{\cos }^2}\theta } \right) \cr
& \frac{{\partial z}}{{\partial r}} = \frac{{2r{{\cos }^2}\theta }}{{{r^2}{{\cos }^2}\theta + 1}} \cr
& \cr
& and \cr
& \cr
& \frac{{\partial z}}{{\partial \theta }} = \frac{\partial }{{\partial \theta }}\left[ {\ln \left( {{r^2}{{\cos }^2}\theta + 1} \right)} \right] \cr
& {\text{Treat }}r{\text{ as a constant}} \cr
& \frac{{\partial z}}{{\partial \theta }} = \frac{1}{{{r^2}{{\cos }^2}\theta + 1}}\frac{\partial }{{\partial \theta }}\left[ {{r^2}{{\cos }^2}\theta + 1} \right] \cr
& \frac{{\partial z}}{{\partial \theta }} = \frac{1}{{{r^2}{{\cos }^2}\theta + 1}}\left( {2{r^2}\cos \theta \left( { - \sin \theta } \right)} \right) \cr
& \frac{{\partial z}}{{\partial \theta }} = \frac{{ - 2{r^2}\cos \theta \sin \theta }}{{{r^2}{{\cos }^2}\theta + 1}} \cr
& \cr
& {\text{Therefore}} \cr
& \frac{{\partial z}}{{\partial r}} = \frac{{2r{{\cos }^2}\theta }}{{{r^2}{{\cos }^2}\theta + 1}}{\text{ and }}\frac{{\partial z}}{{\partial \theta }} = \frac{{ - 2{r^2}\cos \theta \sin \theta }}{{{r^2}{{\cos }^2}\theta + 1}} \cr} $$
