How to prove that
\begin{align} \lim_{(x,y) \rightarrow (0,0)} \frac{x^2-6y^2}{|x| + 3|y|} = 0\ \end{align}
using the Squeeze Theorem? I can work the limit down to $\frac{|x^2 -6y^2|}{|x+y|} $ but can't find a ball $B(x)$ to make the Squeeze Theorem work.