## Thomas' Calculus 13th Edition

$$f=\tan^{-1} (x \space y)+sin^{-1} (yz)+\ln |z|+C$$
We have $$f=\arctan (xy)+g(y,z) \\g(y,z)=\arcsin (yz)+h(z)\\=\tan^{-1} (xy)+sin^{-1} (yz)+h(z)$$ Now, $$f_z=\dfrac{y}{\sqrt {1-y^2z^2}}+\dfrac{1}{z}$$ and $$h(z)=\ln |z|+c$$ Therefore, $$f=\tan^{-1} (x \space y)+\sin^{-1} (yz)+\ln |z|+C$$