Answer
$\ln 3$
Work Step by Step
$ \lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } \ln (\dfrac{3x^2-x^2y^2+3y^2}{x^2+y^2})$
Use polar-coordinates:
$x= r \cos \theta , y = r \sin \theta$ and $r^2=x^2+y^2$
Now, $\lim\limits_{(x,y) \to (0,0) } \ln (\dfrac{3x^2-x^2y^2+3y^2}{x^2+y^2})=\lim\limits_{r \to 0} \ln (\dfrac{ 3r^2 \cos^2 \theta- r^2 \cos^2 \theta r^2 \sin^2 \theta+3r^2 \sin^2 \theta}{r^2})$
So,$\lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{r \to 0} \ln [(3) (\cos^2 \theta + \sin ^2 \theta)]=\ln 3$