I want to calculate the dual group of $(\mathbb{R}\setminus\lbrace 0 \rbrace, \cdot)$.
By assuming $\chi(x)=\exp(2\pi i f(x))$, I concluded from the character property $\chi(xy)=\chi(x)\chi(y)$, that $f(xy)=f(x)+f(y)+k$, $k\in\mathbb{Z}$. Choosing $k=0$ yields $f(x)=c\cdot \log(|x|)$ and all functions of the form $\chi(x)=|x|^{2\pi i \cdot c}=\exp(2\pi i c \log(|x|))$ with $c\in\mathbb{R}$ satisfy the conditions $\chi(xy)=\chi(x)\chi(y)$, $|\chi(x)|=1$ for all $x\in\mathbb{R}\setminus\lbrace 0 \rbrace$ for group characters. So I would naturally identify $\mathbb{R}$ with these characters. My problem now is: Are these all characters of $(\mathbb{R}\setminus\lbrace 0 \rbrace, \cdot)$? If these were all characters, wouldn't $\widehat{\mathbb{R}\setminus\lbrace 0\rbrace}=\mathbb{R}$ contradict $\widehat{\widehat{\mathbb{R}\setminus\lbrace 0\rbrace}}=\mathbb{R}\setminus\lbrace 0\rbrace$, since $\widehat{\mathbb{R}}=\mathbb{R}$?
More thoughts: $z|x|^{2\pi i \cdot c}, z\in\mathbb{C}, |z|=1$ does not generate a character.
It's not clear how you concluded that $f(x)=c\log|x|$. (Regarding your attempt to explain this, see the edit at the bottom.)
The group $\Bbb R\setminus\{0\}$ is the internal direct product of the two subgroups $(0,\infty)$ and $\{1,-1\}$. Since $(0,\infty)$ is isomorphic to the additive group $\Bbb R$, the dual must be isomorphic to $\Bbb R \otimes\Bbb Z_2$. Sorting out the isomorphisms, it turns out the characters are the maps of the form $|t|^{ic}$ and $\text{sgn}(t)|t|^{ic}$ for $c\in\Bbb R$.
Note The rest of this may or may not make sense, given the current version of the question. I've given up trying to keep this answer consistent with the flurry of revisions to the question.
Edit: There's a lot I don't follow about what you wrote, in particular I don't see how "choosing $k$ arbitrarily" leads to that $1/3$. There's a problem somewhere, because $\exp(2\pi i/3)|x|^{2\pi i c}$ is simply not a character.
When you say $f(xy)=f(x)+f(y)+k$ you seem to be assuming that $k$ is independent of $x$ and $y$. This need not be so. If $$f(x)=\begin{cases}\log|x|,&(x>0), \\ \log|x|+\frac12,&(x<0)\end{cases}$$then for every $x$ and $y$ there exists $k\in\Bbb Z$ with $f(xy)=f(x)+f(y)+k$. Seems to me that this leads precisely to the characters that were missing in the first version of your post: If $\chi$ is a character then $\text{sgn}(x)\chi(x)$ is also a character.