Answer
$\dfrac{y}{(1-xy)^2}$ and $\dfrac{x}{(1-xy)^2}$
Work Step by Step
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by treating $y$ as a constant, and vice versa:
$f_x=-(1-xy)^{-2} \times \dfrac{\partial}{\partial x} (1-xy)=\dfrac{y}{(1-xy)^2}$
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $y$, by treating $x$ as a constant, and vice versa:
$f_y=-(1-xy)^{-2} \times \dfrac{\partial }{\partial y} (1-xy)=\dfrac{x}{(1-xy)^2}$
