Answer
$0$
Work Step by Step
$ \lim\limits_{(x,y) \to (0,0) } |f(x,y)|=\lim\limits_{(x,y) \to (0,0) }|xy \times \dfrac{x^2-y^2}{x^2+y^2}|$
or, $=\lim\limits_{(x,y) \to (0,0) } \dfrac{|xy|}{x^2+y^2}|x^2-y^2|$
or, $\lim\limits_{(x,y) \to (0,0) } \dfrac{|xy|}{x^2+y^2}|x^2-y^2| \leq \lim\limits_{(x,y) \to (0,0) } \dfrac{1}{2} \times |x^2-y^2|$
or, $=0$
So, by the Sandwich Theorem$ \lim\limits_{(x,y) \to (0,0) } |f(x,y)|=0$