Answer
$f_x=y^2-3x^2y$
$f_y=2xy-x^3$
Work Step by Step
The function f(x,y)=xy^2-x^3y is provided. We use the definition of partial derivatives as limits in order to find the partial derivatives.
$f_x(x,y)=\lim\limits_{h \to 0} {\frac{(x+h)y^2-(x+h)^3y-xy^2+x^3y}{h}}=\lim\limits_{h \to 0} {\frac{xy^2+hy^2-x^3y-3x^2hy-3xh^2y+h^3y-xy^2+x^3y}{h}}$
Some of the terms cancel leaving the limits:
$\lim\limits_{h \to 0} {\frac{hy^2-3x^2hy-3xhy^2-h^3y}{h}}=\lim\limits_{h \to 0} {y^2-3x^2y-3xhy+h^2y}=[y^2-3x^2y+0+0]=y^2-3x^2y$
Applying a similar procedure for the partial derivative of $y$:
$f_y(x,y)=\lim\limits_{h \to 0}{\frac{x(y+h)^2-x^3(y+h)-xy^2+x^3y}{h}}=\lim\limits_{h \to 0} {2xy+xh-x^3}=[2xy-x^3+0]=2xy-x^3$