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