I'm assuming the first inequality needs absolute values also; if not then exchanging the roles of $x$ and $y$ forces a number and its opposite to both be greater than two.

In this case, there is such an $f$ which is injective:

Let $f(1)=1$.

Assume $f$ has been defined and satisfies the two properties on $\{1,\ldots,m\}$. To define $f(m+1)$ injectively, still satisfying the two above properties, we need that $f(m+1)$ is not in the range of $f$ as defined so far, and also a finite set of inequalities. ($\lfloor\sqrt{m}\rfloor$ of them if $m+1$ is not a square, and an additional $m$ of them if $m+1$ is a square.) If $f(m+1)$ is defined sufficiently large, all these conditions will be met.

Proceed by induction; the resulting function on $\mathbb{N}$ will be as desired.

You ask "What properties should $f$ posess?" If you're looking to contruct an $f$ which grows as slowly as possible, an inspection of the above construction could give one slowish example.

all(x,y)? Your question currently says just one particular pair, which surely can't be what you meant - because then you can't say anything about $f$ except at the three points $\sqrt{x}, x, y$. – Zen Harper Mar 2 '11 at 5:58