Let $(\mathbb{R^{*}},\circ)$ be a semigroup having the property that $\forall x\in \mathbb{R^{*}}, \exists y \in \mathbb{R^{*}}$ such that $x \circ y \neq y \circ x$. Find the functions $f: \mathbb{R^{*}}\to \mathbb{R}$ with $f(x \circ y)=\frac{x+y}{x-y}(f(x)-f(y)), \forall x,y \in \mathbb{R^{*}}, x\neq y.$
What I observed was that $f(x\circ y)=f(y \circ x), \forall x,y \in \mathbb{R^{*}}, x\neq y$. This, together with the other hypothesis, tells us that $f$ is not injective. I don't think that this is really useful, but this is the best I could make out of it.
We also have that $f(x \circ(-x))=0,\forall x\in \mathbb{R^{*}}$ by making $y \to -x$. I am inclined to believe that the only function that satisfies this functional equation is $f=0$, but I cannot make any further progress.
Let $x\in \mathbb{R^{*}}$, choose $ y \in \mathbb{R^{*}}$ such that $x \circ y \neq y \circ x$. If $x \circ y \neq x$ and $y \circ x \neq x$, we have
$$f((x\circ y) \circ x)=\frac{x\circ y +x}{x\circ y-x}(f(x\circ y)-f(x)),$$ $$f(x\circ (y \circ x))=\frac{y\circ x +x}{y\circ x-x}(f(y\circ x)-f(x)),$$ as $f(x\circ y)=f(y\circ x)$, we have $$\left(\frac{x\circ y +x}{x\circ y-x}-\frac{y\circ x +x}{y\circ x-x}\right)(f(x\circ y)-f(x))=0.$$
As $x \circ y \neq y \circ x$ and $g(z)=(z+x)/(z-x)$ is injective, we must have $f(x \circ y)=f(x)$ (Note that this is still true when our assumption is not true). Similarly, we have $f(y \circ x)=f(y)$, $f(x)=f(y)$, and $f(x)=f(x\circ y)=0$.