A Classic Functional Equation

One of the possible solutions of the given functional equation is the function defined by $F(x, y)\equiv 0,$ but in this case, the range of $F$ would contain exactly one number. If there is a pair of real numbers $(a, b),$ such that $F(a, b)=0,$ then we can prove that $F=0$ because $F(x, y)=F(x, a)F(a, y)=F(x, a)F(a,b)F(b,y)=0,$ and again the cardinality of the range of the function $F$ would be one.

So, for a function $F$ to satisfy the two given conditions, $F(x, y) \neq 0$ for all real numbers $x,$ and $y.$ Now, assume that $F$ is different from 0 and satisfies the conditions i) and ii). Then using condition i) with $z=0,$ we obtain that $F(x, y)=\frac{F(x, 0)}{F(y, 0)}\quad\quad (*)$ Using ii) the number of different possible values of $F(x, 0)$ must be one or two. If the number of values of the function $F(x, 0)$ was one then according to (*) the range of $F$ would contain only one number and this would contradict ii). In the case, that $F(x, 0)$ had two values, we have two possibilities: the values are 1 and -1, or at least one of the two values is different from 1 and -1. The first situation would contradict the condition iii) because the function $F(x, y)$ would have only two possible values, 1 or -1. Then let us assume that $F(x, 0)$ has two different values $a$ and $b$ where at least one of them is different from 1 and -1. This also leads to a contradiction because the range of $F$ would contain the set $\{a, b, 1, \frac{a}{b}, \frac{b}{a}\}$ that has at least cardinality 3.

Then there is no function $F$ satisfying the conditions i) and ii). Therefore the answer to the question is $\boxed{0}.$

In the first condition, since we are free to choose any values of x,y,z, we can put x=y to get F(x,x)F(x,z)=F(x,z) => F(x,z)(F(x,x)-1)=0, for all reals x and z. Thus we have 2 cases...
CASE-1: When F(x,z)=0, here condition 1 and 3 are satisfied but not condition 2.
CASE-2: When F(x,x)=1, Condition 3 is not yet completely violated, so we may proceed and see what happens. Using first condition again and putting x=z, we get F(x,y)F(y,x)=F(x,x)=1. Now F(x,y) has to have a value different from 1 to sarisfy condition 3. However if that's the case then F(x,y)=a and F(y,x)=1/a, for non-zero real a, but then if a and 1/a are distinct then condition 2 is violated. Thus a=1/a => a²=1 => a=-1(since we argued earlier a cannot be 1), but now condition 3 is violated.
In both cases, we couldn't get any desirable F(x,y), thus there's no solution to our given conditions which mskes our answer 0.