Solution:

Taking $x=y=1$ in (ii), we get $f(1, z)^2 = f(1, z)$ so that $f(1, z) = 1$ for all $z \neq 0$. Similarly, $x=y=-1$ gives $f(-1, z) = 1$ for all $z \neq 0$. Using the second condition, we also get $f(z, 1) = f(z, -1) = 1$ for all $z \neq 0$.

Observe
$$f\left(\frac{1}{x}, y\right) f(x, y) = f(1, y) = 1 = f(x, 1) = f\left(x, \frac{1}{y}\right) f(x, y)$$
Therefore
$$f\left(x, \frac{1}{y}\right) = f\left(\frac{1}{x}, y\right) = \frac{1}{f(x, y)}$$
for all $x, y \neq 0$.

Now for $x \neq 0,1$, condition (iii) gives
$$1 = f\left(\frac{1}{x}, 1-\frac{1}{x}\right) = f\left(x, \frac{1}{1-\frac{1}{x}}\right)$$
Multiplying by $1 = f(x, 1-x)$, we get
$$1 = f(x, 1-x) f\left(x, \frac{1}{1-\frac{1}{x}}\right) = f\left(x, \frac{1-x}{1-\frac{1}{x}}\right) = f(x, -x)$$
for all $x \neq 0,1$.

But $f(x, -1) = 1$ for all $x \neq 0$ gives
$$f(x, x) = f(x, -x) f(x, -1) = f(x, -x) = 1$$
for all $x \neq 0,1$. Observe $f(1,1) = f(1,-1) = 1$. Hence
$$f(x, x) = f(x, -x) = 1$$
for all $x \neq 0$, which proves (a).

We have
$$1 = f(x y, x y) = f(x, x y) f(y, x y) = f(x, x) f(x, y) f(y, x) f(y, y) = f(x, y) f(y, x)$$
for all $x, y \neq 0$, which proves (b).