#### Answer

$$$ f=\tan^{-1} (x \space y)+sin^{-1} (yz)+\ln |z|+C $$

#### Work Step by Step

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, $$ f=\tan^{-1} (x \space y)+\sin^{-1} (yz)+\ln |z|+C $$