jmc

Setting $x=y=\frac{z}{2}$ in (ii), we get $$f\left(\frac{1}{z}\right)=2f\left(\frac{2}{z}\right)(1)$$for all $z\ne 0.$

Setting $x=y=\frac{1}{z}$ in (iii), we get $$\frac{2}{z}f\left(\frac{2}{z}\right)=\frac{1}{z^2}f{\left(\frac{1}{z}\right)}^2$$for all $z\ne 0.$ Hence, $$2f\left(\frac{2}{z}\right)=\frac{1}{z}f{\left(\frac{1}{z}\right)}^2.(2)$$From (1) and (2), $$f\left(\frac{1}{z}\right)=\frac{1}{z}f{\left(\frac{1}{z}\right)}^2,$$so $$f(x)=xf(x{)}^2(3)$$for all $x\ne 0.$

Suppose $f(a)=0$ for some $a\ne 0.$ Since $f(1)=1,$ $a\ne 1.$ Setting $x=a$ and $y=1-a$ in (iii), we get $$f(1)=a(1-a)f(a)f(1-a)=0,$$contradiction. Therefore, $f(x)\ne 0$ for all $x,$ so from (3), $$f(x)=\frac{1}{x}.$$We can check that this function works, so there is only $\boxed{1}$ solution.