Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 807: 30

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

We need to take the first partial derivatives of the given function. In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa: $f_x=(yz) \times \dfrac{1}{xy} \times (y)=\dfrac{yz}{x}$ $f_y=z \ln (xy) +yz (xy)^{-1} \times (x)=z \ln (xy) +z$ $f_z= \dfrac{\partial f}{\partial z}=y \ln xy$