Handle a function of itself I have a tricky question. I want to handle a function y(x) defined in this way:
$y(x)=f(y(x))g(x)$
Here, $f(y(x))$ and $g(x)$ are smooth functions: is there any way/method to express $y(x)$ as a (complicated) function of $x$ solely?
Maybe it is impossible if $f$ and $g$ are generic.
Edit:
If I have the numerical functions which describe $f(y)$ and $g(x)$, is there any approximated way to express $y(x)$? For example with recursive formulae?
My point is not to guess what form $y(x)$ should have, but deriving it or, at least, inferring it approximately.

Assuming $y(x)$ and $g(u)$ admit inverse functions (which is quite restrictive), we can replace $y(x) = u$:
$$
u = f(u) g\left(y^{-1}(u)\right)
$$
$$
\frac{u}{f(u)} = g\left(y^{-1}(u)\right)
$$
$$
g^{-1}\left(\frac{u}{f(u)}\right) = y^{-1}(u)
$$
Now the left hand side only depends on a free-variable $u$ and on known functions $f$ and $g$. So if you can find $g^{-1}$, then you can find a closed expression for the left hand side as a function of $u$. Call it $w(u)$:
$$
w(u) = g^{-1}\left(\frac{u}{f(u)}\right) = x
$$
Now your $y$ function is given by:
$$
y(x) = w^{-1}(x)
$$
As others pointed out, finding such $y$ is not always possible, and this particular solution is quite restrictive. It will likely only work for cherry picked examples of $f$ and $g$.
