Answer
$\dfrac{\pi}{2}$
Work Step by Step
$ \lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } arctan (\dfrac{|x|+|y|}{x^2+y^2})$
Use polar-coordinates:
$x= r \cos \theta , y = r \sin \theta \\ r^2=x^2+y^2$
Now, $\lim\limits_{(x,y) \to (0,0) } arctan (\dfrac{|x|+|y|}{x^2+y^2})=\lim\limits_{r \to 0} arctan (\dfrac{ |r \cos \theta| +|r \sin \theta|}{r^2})$
and $\lim\limits_{r \to 0} arctan (\dfrac{ |\cos \theta| +|\sin \theta|}{r}) =tan^{-1} (\infty) $
and $\lim\limits_{(x,y) \to (0,0) } f(x,y)=\dfrac{\pi}{2}$
