Let $f: A\to A$ where $A\in (-1,\infty )$ and $f(x + f(y) + xf(y)) = y+f(x)+yf(x)$ for all $x,y \in A$ and $\frac{f(x)}{x} $ is strictly increasing for all $x \in (-1,0) \cup (0,\infty)$.
Then find $f(x)$.
Your options are:-
(A) $xe^{1+x}$
(B) $xlog_{e}(1+x)$
(C) $x \over 1+x$
(D) infinite
How to approach? Need hints for solving.
I don't know what option $D$ means by "infinite," but you can rule out the three other options right away.
Note that if $f(x)=\frac{x}{1+x}$, then $\frac{f(x)}{x}=\frac{1}{1+x}$ is decreasing on $(1,\infty)$. Thus $f(x)$ cannot be $\frac{x}{1+x}$.
Now we rule out $f(x)=xe^{1+x}$ and $f(x)=x\ln(1+x)$. Note that by setting $x=0$, the functional equation requires $$f(f(y))=y+f(0)+yf(0).$$ If $f(x)=xe^{1+x}$, then $f(0)=0$, and thus $$f(f(y))=y,$$ which is not true for, say, $y=1$. Similarly one can rule out $f(x)=x\ln(1+x)$. So by process of elimination I would guess D, but that answer really does not make sense to me at all...