Answer
As a result $f_x=yx^{y-1}$, $f_y=x^y\ln{(x)}$.
Work Step by Step
$f(x,y)=x^y$
In order to find $f_x$ we treat $y$ as a constant and differentiate with respect to $x$.
$f_x=yx^{y-1}$
In order to find $f_y$ we treat $x$ as a constant and differentiate with respect to $y$.
$f_y=x^y\ln{(x)}$
