On the interval $(-1, 1)$, consider the binary operation
$$x*y=\dfrac{2xy+3(x+y)+2}{3xy+2(x+y)+3}$$
with $x, y \in (-1, 1)$. I have to find the number of solutions for the equation:
$$\underbrace{x*x*\ldots*x}_{x\text{ 10 times}}=\frac{1}{10}$$
Finding the left-hand side would be incredibly painful, so I didn't try that. I looked at the previous sub-point of this exercise and it looked like it might help, but I don't know how to use it exactly. In this previous sub-point I showed that for the function:
$$f:(-1, 1) \rightarrow (0, \infty) \hspace{2cm} f(x) = \dfrac{1}{5} \cdot \dfrac{1-x}{1+x}$$
it is true that:
$$f(x * y) = f(x)f(y)$$
$\forall x,y \in (-1, 1)$.
Still I don't know how to find the number of solutions for:
$$\underbrace{x*x*\ldots*x}_{x\text{ 10 times}}=\frac{1}{10}$$
What a strange operation! The trick is to find an isomorphism, and they have given you the hint: because $f$ is a bijection, we can say that the operation of $*$ on $(-1, 1)$ is isomorphic to the operation of usual multiplication on $(0, \infty)$. Thus, in terms of $y = f(x) \in (0, +\infty)$, the problem is simply to solve $y^{10} = f(1/10) \in (0, +\infty)$, which has exactly one solution.
(One should check, quite painlessly, that having an isomorphism is "really this good"; for example, that we also get $x * ... * x = f^{-1}(f(x)^n)$ and so on. Indeed, a bijection satisfying the homomorphism law is the correct notion of an isomorphism between magmas.)