#### Answer

$$\dfrac{yz}{x}; \\z \ln (xy) +z ; \\y \ln (xy)$$

#### Work Step by Step

Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by treating $y$ and $z$ as a constant:
$f_x=y (yz) \times \dfrac{1}{xy} =\dfrac{yz}{x}$
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $y$, by treating $x$ and $z$ as a constant:
$f_y=x z \times \ln (xy) +yz \times (xy)^{-1} =z \ln (xy) +z$
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $z$, by treating $x$ and $y$ as a constant:
$f_z= \dfrac{\partial f(x,y)}{\partial z}=y \ln xy$