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