Answer
Yes
Work Step by Step
Given : $1 - \dfrac{x^2 y^2}{3} \lt \dfrac{\tan^{-1} (xy)}{xy} \lt 1$
Since, $\lim\limits_{(x,y) \to (0,0) } (1 - \dfrac{x^2 y^2}{3})=1$
and $\lim\limits_{(x,y) \to (0,0) } 1=1$
This implies that the limit for $\lim\limits_{(x,y) \to (0,0) } (\dfrac{\tan^{-1} (xy)}{xy} )=1$ by the Sandwich Theorem.
So, our answer is Yes.