Can you find function which satisfies $f(ab)=\frac{f(a)}{f(b)}$? Can you find function which satisfies $f(ab)=\frac{f(a)}{f(b)}$? For example $log(x)$ satisfies condition $f(ab)=f(a)+f(b)$ and $x^2$ satisfies $f(ab)=f(a)f(b)$?
 A: You could take a relatively trivial function like $f(x) = 1$. Or a slightly more general version that takes everything to the identity.
A: Assuming the function is defined on non-zero real numbers, and takes all non-zero values (but please do see below for a generalization), one has first
$$
f(1) = f(1 \cdot 1) = \frac{f(1)}{f(1)} = 1,
$$
and then for all $x$
$$
f(x) = f( 1 \cdot x) = \frac{1}{f(x)},
$$
so that $f(x) \in \{1, -1 \}$. Then
$$
f(x y) = \frac{f(x)}{f(y)} = f(x) f(y),
$$
so that we get that
$$
f(x) = 1, \qquad\text{or}\qquad f(x) = \operatorname{sgn}(x).
$$

Addendum One may consider the same problem for $f: G \to H$, where $G, H$ are groups (multiplicatively written, with identity $1$), and then it is possibly clearer. (See also the comments to OP.)
The condition is now $f(ab) = f(a) f(b)^{-1}$. Once more, $f(1) = 1$, and $f(x) = f(1 \cdot x) = f(1) f(x)^{-1} = f(x)^{-1}$, so all values of $f$ are involutions (or the identity) and $f$ is a group homomorphism.
So in this case we have that $f$ is a morphism of $G$ onto a(n abelian) subgroup of $H$ whose non-identity elements are involutions. (Clearly there is a non-trivial such $f$ if and only if $G$ has a non-trivial quotient of exponent $2$.)

A: Let us reformulate the question as classify all maps $f : G \rightarrow H$ which need not be group morphism that satisfies the condition $f(ab)=f(a)f(b)^{-1}$
A simple calculation shows that $f(e)=f(x)f(x^{-1})^{-1}= f(x^{-1})f(x)^{-1}$ or we have $f(x)=f(e)^{-1}f(x^{-1})=f(e)f(x^{-1})$ or $f(e)^{-1}=f(e)$ Now $f(x)=f(ex)=f(e)f(x)^{-1}=f(e)^{-1}f(x)^{-1}=(f(x)f(e))^{-1}=(f(x)f(e)^{-1})^{-1}=f(xe)^{-1}=f(x)^{-1}$ so even if we don't assume a group morphism we have the image involutive. And hence it's a group morphism $f(ab)=f(a)f(b)^{-1}=f(a)f(b)$
