, 2015

Put $x=y=0$ into equation (b) to get $f(0)=0$. Put $y=-x$ into equation (b) to get $f(-x)=-f(x)$ and deduce that $f(x-y)=f(x)-f(y)$.

Let $x \in \mathbb{R} \setminus \{0,1\}$. We have
$$f\left(\frac{1}{x-1}\right)=\frac{f(x-1)}{(x-1)^2}=\frac{f(x)-1}{(x-1)^2}$$
On the other hand,
$$\begin{aligned} f\left(\frac{1}{x-1}\right) & = f\left(\frac{1}{x-1}+1\right)-1 = f\left(\frac{x}{x-1}\right)-1 = \frac{f\left(\frac{x-1}{x}\right)}{\left(\frac{x-1}{x}\right)^2}-1 \\ & = \frac{1-f\left(\frac{1}{x}\right)}{\left(\frac{x-1}{x}\right)^2}-1 = \frac{x^2-f(x)}{(x-1)^2}-1 = \frac{2x-1-f(x)}{(x-1)^2} \end{aligned}$$
We deduce from these two relations that for all $x \in \mathbb{R} \setminus \{0,1\}$, we have $f(x)-1=2x-1-f(x)$ which is equivalent to $f(x)=x$. Therefore, $f(x)=x$ for all $x \in \mathbb{R}$.

Conversely, this function satisfies all the conditions of the problem.