Find all real-valued functions $f$ defined on the set of all non-zero real numbers such that:

(i) $f(1)=1$,

(ii) $f\left(\frac{1}{x+y}\right)=f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right)$ for all non-zero $x, y, x+y$,

(iii) $(x+y) f(x+y)=x y f(x) f(y)$ for all non-zero $x, y, x+y$.
Substituting $x=y=\frac{1}{2} z$ in (ii) we get

$$
f\left(\frac{1}{z}\right)=2 f\left(\frac{2}{z}\right)
$$

for all $z \neq 0$. Substituting $x=y=\frac{1}{z}$ in (iii) yields

$$
\frac{2}{z} f\left(\frac{2}{z}\right)=\frac{1}{z^{2}}\left(f\left(\frac{1}{z}\right)\right)^{2}
$$

for all $z \neq 0$, and hence

$$
2 f\left(\frac{2}{z}\right)=\frac{1}{z}\left(f\left(\frac{1}{z}\right)\right)^{2} .
$$

From (1) and (2) we get

$$
f\left(\frac{1}{z}\right)=\frac{1}{z}\left(f\left(\frac{1}{z}\right)\right)^{2},
$$

or, equivalently,

$$
f(x)=x(f(x))^{2}
$$

for all $x \neq 0$. If $f(x)=0$ for some $x$, then by (iii) we would have

$$
f(1)=(x+(1-x)) f(x+(1-x))=(1-x) f(x) f(1-x)=0
$$

which contradicts the condition (i). Hence $f(x) \neq 0$ for all $x$, and (3) implies $x f(x)=1$ for all $x$, and thus $f(x)=\frac{1}{x}$. It is easily verified that this function satisfies the given conditions.