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?

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.)
