Find all pairs of functions $ f,g : \mathbb{R}\to \mathbb{R}$ such that
(a) if $ x < y$, then $ f(x) < f(y)$;
(b) for all $ x,y \in \mathbb{R}$, $ f(xy) = g(y)f(x) + f(y)$.
My work :
Let $ P(x,y) : f(xy) = g(y)f(x) + f(y)$
$x<y \iff f(x) < f(y)$
$ P(x,1) : f(x)\big(1-g(1)\big) = f(1)$
$g(1) = 1 \implies f(xy) = f(x) + f(y)$
so $f(1) = f(1) + f(1) \implies f(1) = 0$
$ P(x,1) :$
$f(1) = 0 \implies f(x) = g(1)f(x)f(1)$
so $f(x)\big(1-g(1)\big) = 0 \implies g(1) = 1$
Please suggest how to proceed.
The answer is, $(f,g)$ satisfies $g=\text{sgn}(x)\exp({b\log|x|})$ and $f(x)=a(g(x)-1)$ for some real numbers $a>0$ and $b>0$.
It is easy to see that all such $(f,g)$ satisfy the conditions. We will show that they are the only possibilities.
The proof is the following:
Take $y=1$, we get $$f(x)=g(1)f(x)+f(1).$$ Since $f$ is not constant, $f(1)=0$, $g(1)=1$.
Take $y=0$, $$ f(0)=g(0)f(x)+f(0). $$ Hence $g(0)=0$.
Take $x=0$ $$ f(0)=g(y)f(0)+f(y), $$ Hence $$f(x)=a(g(x)-1),$$ here we assume $f(0)=-a.$ Of course $a>-f(1)=0$ by assumption, and $g$ is monotonic increasing as $f$ is.
Use this equation to the original one, we get $$ g(xy)=g(x)g(y). $$ Take $x=y=-1$, note that $g$ is monotonic, hence $g(-1)=1$. Take $y=-1$, we know that $g$ is an odd function. It is enough to show that $g=\exp({b\log x})$ for $x>0$. This is actually a standard Cauchy functional equation problem.
Take $G(x)=\log (g(e^x))$. Note that it is well defined because $g(y)>0$ if $y>0$. Also note that $G$ is monotonic increasing and $$ G(x+y)=G(x)+G(y). $$ Hence $G(x)=bx$ for some $b>0$ and we finished the proof.