Answer
$ a.\quad \displaystyle \frac{\partial z}{\partial x}=g(y)\cdot f'(x), \quad \displaystyle \frac{\partial z}{\partial y}=f(x)g'(y).$
$b.\quad \displaystyle \frac{\partial z}{\partial x}=yf'(xy), \quad \displaystyle \frac{\partial z}{\partial y}=xf'(xy)$.
$c.\quad \displaystyle \frac{\partial z}{\partial x}=\frac{f'(\frac{x}{y})}{y}, \quad \displaystyle \frac{\partial z}{\partial y}=-\frac{xf'(\frac{x}{y})}{y^{2}}$.
Work Step by Step
$a.$
$z =f(x)g(y)$
For $\displaystyle \frac{\partial z}{\partial x}$, y and g(y) are constant,
$\displaystyle \frac{\partial z}{\partial x}=g(y)\cdot f'(x)$
For $\displaystyle \frac{\partial z}{\partial y}$, x and f(x) are constant,
$\displaystyle \frac{\partial z}{\partial y}=f(x)g'(y)$
$b.$
$z=f(xy),$ a composition. Chain rule applies.
Let $u=xy$. Then $\displaystyle \frac{\partial u}{\partial x}=y$ and $\displaystyle \frac{\partial u}{\partial y}=x$.
$\displaystyle \frac{\partial z}{\partial x}=\frac{df}{du}\frac{\partial u}{\partial x}=\frac{df}{du}\cdot y=yf'(u)=yf'(xy)$
$\displaystyle \frac{\partial z}{\partial y}=\frac{df}{du}\frac{\partial u}{\partial y}=\frac{df}{du}\cdot x=xf'(u)=xf'(xy)$.
$c.$
$z=f(\displaystyle \frac{x}{y}),$ a composition. Chain rule applies.
Let $u=\displaystyle \frac{x}{y}$. Then $\displaystyle \frac{\partial u}{\partial x}=\frac{1}{y}$ and $\displaystyle \frac{\partial u}{\partial y}=-\frac{x}{y^{2}}$.
$\displaystyle \frac{\partial z}{\partial x}=\frac{df}{du}\frac{\partial u}{\partial x}=f'(u)\frac{1}{y}=\frac{f'(\frac{x}{y})}{y}$
$\displaystyle \frac{\partial z}{\partial y}=\frac{df}{du}\frac{\partial u}{\partial y}=f'(u)(-\frac{x}{y^{2}})=-\frac{xf'(\frac{x}{y})}{y^{2}}$.
