## Thomas' Calculus 13th Edition

$$f=y \space x \sin (z)+c$$
We have $$f=y \space x \sin (z) +g(y,z)$$ and $$g(y,z)=0 \implies g(y,z)=h(z)=0$$ So, $f=y \space x \sin z+h(z)$ Now, $f_z=yx \cos z+h'(z)=yx \cos z$ So, $$f=y \space x \sin (z)+c$$