Answer
$f=yx \sin z+c$
Work Step by Step
Here, $f=yx \sin z+g(y,z)$
and $g(y,z)=0$ or, $g(y,z)=h(z)=0$
Then $f=yx \sin z+h(z)$
This implies that
$f_z=yx \cos z+h'(z)=yx \cos z$
Hence, $f=yx \sin z+c$
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