Let $g(x) = f(\frac{1}{x})$. We quickly see using (ii) that $g$ is an additive function, so $g(x) = x$ at all points except $x = 0$. This means $\boxed{f(x) = \frac{1}{x}}$ for all $x \not = 0$. We easily see this function satisfies (iii), and so we are done.

dinoboy wrote: We quickly see using (ii) that $g$ is an additive function, so $g(x) = x$ at all points except $x = 0$. It is wrong that all additive functions are linear functions. I have seen that you had jumped too quickly into this conclusion in this thread as well. To be clear, the solutions to Cauchy's functional equation have a lot to do with the Hamel basis of the reals. The condition (iii) is given so that other "complicated" functions are eradicated.

Full solution can be seen here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=618322&p=3687134#p3687134

Let $f(x)=\frac{1}{g(x)}$, then from (iii), \[\frac{g(x+y)}{x+y}=\frac{g(x)}{x}\cdot \frac{g(y)}{y}\]let $h(x)=\frac{g(x)}{x}$, $h$ is multiplicative, then letting $y=1$ and since $h(1)=1$, we have \[h(1+x)=h(x)\ \ \forall x\ne 0.\]Since $h(x)$ is even and a constant function, we can get that $f(x)=\frac{1}{x} \ \ \forall x\in \mathbb{Z}, x\ne 0$, now let $x=\frac{1}{u}, y=\frac{1}{v}, u,v\in \mathbb{Z} \setminus\{0\}$ and $u+v\ne 0$, from (ii), \[f(\frac{uv}{u+v})=f(u)+f(v)=\frac{u+v}{uv}\]and so we have $f(x)=\frac{1}{x}$ and it's easy to see that this satisfies the equation. $\blacksquare$