<s>[INST] 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?
 [/INST] $$\frac{x^2+2xy+y^2}{x^2-y^2}=\frac{(x+y)^2}{(x-y)(x+y)}=\frac{x+y}{x-y} \ge (x+y) \implies \frac{1}{x-y}\ge 1$$
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.
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