Answer
$f_{y}(x, y, z)=\displaystyle \frac{x+z}{(x+y+z)^{2}},\qquad f_{y}(2,1, -1)=\displaystyle \frac{1}{4}$.
Work Step by Step
Treat x and z as constant to calculate $f_{y}(x, y ,z)$. Apply the quotient rule.
$f_{y}(x, y, z) =\displaystyle \frac{[\frac{\partial}{\partial y}(y)](x+y+z)-y[\frac{\partial}{\partial y}(x+y+z)]}{(x+y+z)^{2}}$
$=\displaystyle \frac{1(x+y+z)-y(1)}{(x+y+z)^{2}}$
$=\displaystyle \frac{x+z}{(x+y+z)^{2}}$
$f_{y}(2,1, -1)=\displaystyle \frac{2+(-1)}{(2+1+(-1))^{2}}=\frac{1}{4}$.
