Solve $\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$ 
Find the set of integer solutions $(x,y)$ to
  $$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$$

I can't seem to multiply both sides by the expression in the denominator. 
Nor can I simplify and cancel any terms. How should I it?
 A: Hint
$$\frac{x^2+2xy+y^2}{x^2-y^2}=\frac{(x+y)^2}{(x+y)(x-y)}=\frac{x+y}{x-y}$$
Edit
So, the initial inequality is equivalent to
$$\frac{x+y}{x-y}>x+y.$$
If $x+y>0$ it is 
$$\frac{1}{x-y}>1,$$ which has no solution. (If $x-y<0$ then $\frac{1}{x-y}<0$ can't be greater than $1.$ If $x-y>0$ then $\frac{1}{x-y}\le 1$ can't be greater than $1.$)
Thus, it must be $x+y<0.$ In such a case, the inequality is equivalent to 
$$\frac{1}{x-y}<1,$$ which is satisfied if $x-y<0$ or $x-y>1.$ 
So we have the regions $x+y<0, x-y<0$ and $x+y<0,x-y>1.$ 
Thus, the solution is $$\{(x,y)\in \mathbb{Z}^2:x< y<-x \:\mathrm{or} \: y+1<x<-y\}.$$
(Remark for readers just coming to this post:  All the intial comments below this answer, down to the one with a little boldface, pertain to earlier versions of the answer.  There's no need to read them.)
A: $$\frac{x^2+2xy+y^2}{x^2-y^2}=\frac{(x+y)^2}{(x-y)(x+y)}=\frac{x+y}{x-y} \ge (x+y) \implies \frac{1}{x-y}\ge 1$$
Without loss of generality, assume that $x \ge y$, and fiddle around with this expression.

Edit: We do the standard LHS-RHS trick:
$$\frac{x^2+2xy+y^2}{x^2-y^2}-(x+y)>0 \\ LHS= \frac{(x+y)^2}{(x-y)(x+y))}-(x+y) = \frac{(x+y)^2-(x-y)(x+y)^2}{(x^2-y^2)}$$ Hence, $$\frac{(x+y)^2(1-(x-y))}{x^2-y^2}>0$$
Then I think from here take cases regarding whether $x^2-y^2>0 \text{ or } x^2-y^2<0$ 
, and solve the inequation.
A: $$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y \iff \frac{x+y}{x-y}> x+y$$
Case 1: $x+y > 0$
In this case, we have $\dfrac1{x-y}> 1 \iff 0 < x-y< 1$ which is not possible.
Case 2: $x+y < 0$
In this case, we have $\dfrac1{x-y} < 1 \iff x-y < 0$ or $x-y > 1$.   
Now $x+y < 0, \; x-y < 0$ gives $x < 0, \;x < y < -x$
and $x+y < 0, \; x-y > 1$ gives $y < 0, \;1+y< x < -y$ 
Hence we can write the possible solutions also as integer pairs satisfying $\{|y| < -x\} \cup \{|x| < -y-\frac12\}$.
A: Solving algebraic inequalities can be a tricky business, because the urge to cancel can be overwhelming.  There are times when it's OK to cancel, and there are times when it's not.  I for one almost always make mistakes, so I've learned to proceed super-cautiously (and even so, it's a lesson I have to learn over and over again).  So let me do things very slowly.
The first step in solving
$${x^2+2xy+y^2\over x^2-y^2}\gt x+y$$
is to rewrite it as
$${(x+y)^2\over(x+y)(x-y)}\gt x+y$$
The first urge is to cancel an $(x+y)$ from the numerator and denominator on the left hand side, leaving
$${x+y\over x-y}\gt x+y$$
Even this should be done with caution, because technically the left hand side of the original inequality is not defined when $x+y=0$. If the inequality had a $\ge$ sign instead of the $\gt$ sign, this would be a problem.  Fortunately, $0\gt0$ does not hold, so points on the line $x+y=0$ are ruled out anyway, so that cancellation really is OK.
The next urge is to cancel the $x+y$ from the two sides of the inequality, leaving
$${1\over x-y}\gt1$$
This is the urge to resist.  The reason, when you stop to think about it (and therein lies the rub -- it's all too easy not to stop and think), is obvious:  You don't know whether $x+y$ is positive or negative, and inequalities reverse direction when you cancel a negative number.  We all "know" this, but many of us tend to forget it at inopportune moments.
There are various correct ways to proceed.  One, as illustrated in Macavity's answer, is to split things into two cases:  $x+y\gt0$ and $x+y\lt0$.  Another way is to move everything to one side of the inequality (which is always OK), obtaining, in this case,
$${(x+y)(1-(x-y))\over x-y}\gt0$$
My personal preference here is to rewrite this as
$$(x+y)(x-y)(1-(x-y))\gt0$$
My reason for doing this is that it puts all three factors on an equal basis; my justification is that all I've done is multiplied both sides of the inequality by the positive quantity $(x-y)^2$.
At this point I find it convenient to graph the three lines $x+y=0$, $x-y=0$, and $x-y=1$, and use the fact that the sign of the product changes every time you cross one of these lines.  If you do this, you find that the set of real solutions is the interior of the three gray regions in

where the bounding lines are $x+y=0$, $x-y=0$, and $x-y=1$.  (Sorry for the crummy picture; it was just easy to gin up.)  However, the region shaded in light gray contains no integer points, so that leaves the two dark gray regions.  
The final question is how to describe these regions correctly.  Macavity gave one way.  Here's another:
$$\{(x,y):x\lt y\lt-x\}\cup\{(x,y):y+1\lt x\lt-y\}$$
The first set corresponds to the triangle at opens to the left; the second to the one that opens to the bottom.  
I'm sorry if this seems like an awful lot of exposition for a "simple" inequality.  If all you want is a succinct solution, Macavity's is the way to go.  But given that the OP accepted an answer that was (and, at the time I'm posting this, still is) incorrect, I thought it would be worth belaboring a point or two.  I don't mean that as a criticism, either of the OP or the person who gave the incorrect answer.  Believe me, any mistakes they've made pale in comparison to my own.
