Hard functional equation: $ f \big ( x y + f ( x ) \big) = f \big( f ( x ) f ( y ) \big) + x $
Exchanging $x$ and $y$ and substracting, it follows $f(xy+f(x))-f(xy+f(y))=x-y$. In particular, if $f(x)=f(y)$ then $x=y$.
The equation also tells us that if $r > f(x)$, we can find a $y> 0$ such that $r=f(x)+xy$, so $f(r)=f(xy+f(x))=f(f(x)f(y))+x > x$, ie that if $r > f(x)$, $f(r) > x$.
In particular, if $x > f(x)$, $f(x) > x$, so we have, for all $x$, $f(x) \geq x$.
Now, let us fix some $x > 0$ such that $f(x)>x$.
Define, for any $y > 0$, $g(y)=\frac{f(x)}{x}(f(y)-1)$. If $g(y)>0$, then note that $xg(y)+f(x)=f(x)f(y)$, thus $f(xy+f(x))=f(xg(y)+f(x))+x$.
Therefore, if $y >0$ and $g^n(y)>0$ is defined, $0<f(xg^n(y)+f(x))=f(xy+f(x))-nx$. As a consequence, $n < \frac{f(xy+f(x))}{x}+1$ (the precise estimate is irrelevant, just remember tha the RHS is explicit in $x$ and $y$).
In particular, there exists some $n \geq 0$ (depending on $x,y$) such that $g^n(y) > 0$ is defined and $g^{n+1}(y) \leq 0$.
Now, take $y > \alpha$, where $f(x)(\alpha-1)=x\alpha$. Then $g(y)=\frac{f(x)}{x}(f(y)-1) \geq \frac{f(x)}{x}(y-1) > f(x)(\alpha-1)/x=\alpha$.
We find that $g^n(y)$ is defined and positive for all $n$, a contradiction.
$$f(xy + f(x)) = f(f(x)\cdot f(y)) + x.$$
Let us substitute $y = 1$:
$$f(x + f(x)) = f(f(x)\cdot f(1)) + x.$$
Now, let us substitute into the initial equation $x = 1$:
$$f(y + f(1)) = f(f(1)\cdot f(y)) + 1.$$
In the latter equation, let us replace $y$ by $x$:
$$f(x + f(1)) = f(f(x)\cdot f(1)) + 1.$$
Now, we have
$$ f(x + f(x)) = f(f(x)\cdot f(1)) + x \\ f(x + f(1)) = f(f(x)\cdot f(1)) + 1 $$
Let $g(x) = f(f(x)\cdot f(1))$. Then, we have
$$ f(x + f(x)) = g(x) + x \\ f(x + f(1)) = g(x) + 1 $$
We see that a linear shift in the argument of the function $f(x)$ results in a linear shift in the values of function g(x).
- Shifting by $f(x)$, i.e. $f(x + f(x))$ implies shifting by $x$.
- Shifting by $f(1)$, i.e. $f(x + f(1))$ implies shifting by $1$.
This is true if both $f(x)$ and $g(x)$ are linear functions, particularly if $$f(x) = x.$$
Let us check that $f(x) = x$ is a solution:
$$f(xy + f(x)) = f(f(x)\cdot f(y)) + x \Leftrightarrow f(xy + f(x)) = xy + x \text{ and }f(f(x)f(y)) + x = xy + x \text{ (TRUE). }$$