Answer
$$\frac{\partial}{\partial x}f(x,y)=yx^{y-1}$$
$$\frac{\partial}{\partial y}f(x,y)=x^y\ln x$$
Work Step by Step
Regard $y$ as constant and differentiate with respect to $x$. Then $f$ is treated as a simpe power function:
$$\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}x^y=yx^{y-1}.$$
Regard $x$ as constrant and differentiate with respect to $y$. Then $f$ is treated as a simpe exponential function:
$$\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}x^y=x^y\ln x.$$
