Solve over the positive real numbers the functional equation  
$$ f(x y+f(x))=x f(y)+2 $$
$ The answer is $f(x) \equiv x+1$, which is easily verified to be the only linear solution. We show conversely that $f$ is linear. Let $P(x, y)$ be the assertion. Claim $-f$ is weakly increasing. Claim (Up to an error of $2, f$ is linear) - We have $$ |f(x)-(K x+C)| \leq 2 $$ where $K:=\frac{2}{f(1)}$ and $C:=f(f(1))-2$ are constants. $$ 2\left\lfloor\frac{x}{f(1)}\right\rfloor+C \leq f(x) \leq 2\left\lceil\frac{x}{f(1)}\right\rceil+C $$ which implies the result. Rewrite the previous claim to the simpler $f(x)=K x+O(1)$. Then for any $x$ and $y$, the above claim gives $$ K(x y+K x+O(1))+O(1)=x f(y)+2 $$ which means that $$ x \cdot\left(K y+K^{2}-f(y)\right)=O(1) . $$ If we fix $y$ and consider large $x$, we see this can only happen if $K y+K^{2}-f(y)=0$, i.e. $f$ is linear.