Answer
Limit does not exist.
Work Step by Step
$ \lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } \dfrac{x^2-y^2}{x^2+y^2}$
Use polar-coordinates:
$x= r \cos \theta , y = r \sin \theta$ and $r^2=x^2+y^2$
$\lim\limits_{(x,y) \to (0,0) } \dfrac{x^2-y^2}{x^2+y^2}=\lim\limits_{r \to 0} \cos (\dfrac{r^2 \cos^2 \theta- r^2 \sin ^2 \theta}{r^2})$
and $\lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{r \to 0} \cos (\cos^2 \theta-\sin ^2 \theta)=\cos^2 \theta-\sin ^2 \theta$
We see that $\cos^2 \theta-\sin ^2 \theta$ is not a unique value.
Thus, $ \lim\limits_{(x,y) \to (0,0) } f(x,y)=DNE$
