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} $
 A: I always like to let
variables go to zero.
So,
if we let
$x = u+1$
and
$y = v+1$,
$\begin{array}\\
\frac {x^2+2xy-3y^2}{x^2-y^2}
&=\frac {(u+1)^2+2(u+1)(v+1)-3(v+1)^2}{(u+1)^2-(v+1)^2}\\
&=\frac {u^2+2u+1+2(uv+u+v+1)-3(v^2+2v+1)}{(u^2+2u+1)-(v^2+2v+1)}\\
&=\frac {u^2+4u+2uv-3v^2-4v)}{u^2+2u-v^2-2v}\\
&=\frac {u^2+2uv-3v^2+4u-4v}{u^2-v^2+2u-2v}\\
&=\frac {(u-v)(u+3v)+4(u-v)}{(u-v)(u+v)+2(u-v)}\\
&=\frac {(u+3v)+4}{(u+v)+2}
\quad\text{(for }u \ne v)\\
\end{array}
$
The limit is,
therefore,
$\lim_{u \to 0, v \to 0} \frac {(u+3v)+4}{(u+v)+2}
=\frac{4}{2}
=2
$.
In this case,
it was harder.
A: $x^2+2xy - 3y^2=(x+3y)(x-y), x^2-y^2 = (x-y)(x+y)\to L = \displaystyle \lim_{(x,y) \to (0,0)} \dfrac{x+3y}{x+y}$. The limit varies so it doesn't exist.... 
Update: Based on your recent edit, the new limit is $2$ since now $(x,y) \to (1,1)$.
A: $$ \lim_{(x,y)\to(1,1)} \frac {x^2+2xy-3y^2}{x^2-y^2}$$
factorizing numerator and denominator we have,
$$ \lim_{(x,y)\to(1,1)} \frac {(x+3y)(x-y)}{(x+y)(x-y)}$$
$$\implies  \lim_{(x,y)\to(1,1)} \frac {x+3y}{x+y} $$
$$=\frac{4}{2}=2$$
A: Hint: As in many limit problems about fractions, factorize both the numerator and the denominator and cancel any identical terms. The denominator is the difference of two squares, so is easy to factor. The numerator is a trinomial and is not much more difficult.
A: First you factor out the $(x-y)$ term to get
$$
f(x,y) = \frac{x+3y}{x+y}.
$$
To simplify further let $x=u+1$ and $y=v+1$:
$$
g(u,v) = \frac{4+u+3v}{2+u+v}.
$$
and we look for the limit as $(u,v) \rightarrow (0,0)$.
Given $\epsilon\gt0 $ suppose $\|(u,v)\|\lt \epsilon$. We have $|u|\lt \epsilon$ and $|v|\lt \epsilon$ and we calculate
$$
|g(u,v)-2|=\frac{|v-u|}{|2+u+v|}
$$
The numerator is bounded above by $|u|+|v| \le 2 \epsilon$.
The denominator is bounded below by $2-|u|-|v| \gt 2-2 \epsilon \gt 1$ for $\epsilon \lt{1\over2}$
So the whole thing is bounded above by $2\epsilon$. This is your $\delta$.
