Chapter 15 - Section 15.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises - Page 849: 12

$\tan^{-1} (xy)+sin^{-1} (yz)+\ln |z|+c$

Here, $f=\tan^{-1} (xy)+g(y,z)$ and $g(y,z)=\sin^{-1} (yz)+h(z)$ Then $f=\tan^{-1} (xy)+sin^{-1} (yz)+h(z)$ This implies that $f_z=\dfrac{y}{\sqrt {1-y^2z^2}}+\dfrac{1}{z}$ $h(z)=\ln |z|+c$ Hence, $f=\tan^{-1} (xy)+sin^{-1} (yz)+\ln |z|+c$