Prove using $ \varepsilon-\delta $ that $ \lim_{(x,y)\to(1,1)} \frac {x^2+2xy-3y^2}{x^2-y^2} = 2 $ 
Prove limit using $ \varepsilon-\delta $ definition that: $$ \lim_{(x,y)\to(1,1)} \frac
 {x^2+2xy-3y^2}{x^2-y^2} = 2 $$

I've been reading quite a lot about how to prove limits; so I want to show what I've done so far in this one, so you can tell me any suggestion (tricks) and even point out any mistake.
What I've tried:
I want to find $ \delta $ for every $ \varepsilon $ that verifies: $$ 0 < \left \| (x,y) - (1,1) \right \| < \delta \Rightarrow \left | \frac {x^2+2xy-3y^2}{x^2-y^2} - 2 \right | < \varepsilon $$
So here is my attempt:
$$ \begin{align*}
\left | \frac {x^2+2xy-3y^2}{x^2-y^2} - 2 \right | &\overset{(1)}{=} \left | \frac {y-x}{y+x} \right | \\ 
 &\overset{(2)}{\leq} \frac {|x|+|y|}{|x+y|} \\ 
 &\overset{(3)}{\leq} \frac {2\left \| (x,y)-(1,1) \right \|}{|x+y|} \\
 &\overset{(4)}{\leq} 4\left \| (x,y)-(1,1) \right \| \\
 & < \ 4 \delta
\end{align*} $$
So I can take $ \delta = \varepsilon / 4 $. Is this right?
Justifications:
(1) Basic operations.
(2) Triangle inequality.
(3) I used $ |x| \leq \left \| (x,y)-(1,1) \right \| $.
(4) I supposed $ |x| < 1/2 $ and also $ |y| < 1/2 $ then $ |x+y| < 1/2 $. (I don't understand why this step holds though).
(5) The metric I'm using is: $ \left \| (x,y) \right \| = \sqrt{x^2 + y^2} $

$$ \lim_{(x,y)\to(1,1)} \frac {x^2+2xy-3y^2}{x^2-y^2}$$
factorizing numerator and denominator we have,
$$ \lim_{(x,y)\to(1,1)} \frac {(x+3y)(x-y)}{(x+y)(x-y)}$$
$$\implies  \lim_{(x,y)\to(1,1)} \frac {x+3y}{x+y} $$
$$=\frac{4}{2}=2$$
