Answer
$\dfrac{1}{2}$
Work Step by Step
Consider $l=\lim\limits_{(x,y) \to (1,1)} \dfrac{x-y}{x^2-y^2}$
and $l=\lim\limits_{(x,y) \to (1,1)} \dfrac{x-y}{(x-y)(x+y)}$
Hence, $l=\lim\limits_{(x,y) \to (1,1)} \dfrac{1}{(x+y)}=\dfrac{1}{(1+1)}=\dfrac{1}{2}$
