Answer
$a.\quad \displaystyle \frac{\partial z}{\partial x}=f'(x),\quad \frac{\partial z}{\partial y}=g'(y)$.
$b.\quad \displaystyle \frac{\partial z}{\partial x}=f'(x+y),\quad \frac{\partial z}{\partial y}=f'(x+y)$.
Work Step by Step
$a.$
$z =f(x)+g(y)$
For $\displaystyle \frac{\partial z}{\partial x}$, y and g(y) are constant,
$\displaystyle \frac{\partial z}{\partial x}=\frac{\partial}{\partial x}[f(x)]+0=f'(x)$,
For $\displaystyle \frac{\partial z}{\partial y}$, x and f(x) are constant,
$\displaystyle \frac{\partial z}{\partial y}=0+\frac{\partial}{\partial y}[g(x)]=g'(y)$
$b.$
$z =f(x+y)$, a composition. Chain rule applies.
Let $u=x+y$. Then$, \displaystyle \frac{\partial u}{\partial x}=1, \quad \frac{\partial u}{\partial y}=1$
$\displaystyle \frac{\partial z}{\partial x}=\frac{df}{du}\frac{\partial u}{\partial x}$
$\displaystyle \frac{\partial z}{\partial x}=\frac{df}{du}(1)=f'(u)$
$=f'(x+y)$
$ \displaystyle \frac{\partial z}{\partial y}=\frac{df}{du}\frac{\partial u}{\partial y}=\frac{df}{du}(1)=f'(u)$
$=f'(x+y)$.
