Find all functions $ g : \mathbb R \to \mathbb R $ with the property:
There exists a strictly monotonic function $ f : \mathbb R \to \mathbb R $ such that $$ f ( x + y ) = f ( x ) g ( y ) + f ( y ) , \forall x , y \in \mathbb R $$
We have one official solution, which shows that $ g ( x ) $ must be of the form $ e ^ { k x } $, or simply $ a ^ x $, $ a > 0 $. I'm looking for other solutions, just trying to get some new ideas. If non is posted, I'll post the solution I have.
If any $f(x) = 0$ or $g(y) = 0$, then your equation says $f(x+y) = f(y)$. Since $f$ is supposed to be strictly monotone, this can only happen if $x = 0$, and in particular $g$ is never $0$. Conversely, substituting $x=0$ in your equation we see that $f(0) = 0$.
Now according to your equation, if $x \ne 0$ $$ g(y) = \dfrac{f(x+y)-f(y)}{f(x)} $$ which must not depend on $x$.
Now a monotone function is almost everywhere differentiable. If $f$ is differentiable at $y$,
$$f'(y) = \lim_{h \to 0} \dfrac{f(h+y) - f(y)}{h} = g(y) \lim_{h \to 0} \dfrac{f(h) - f(0)}{h}$$
so (recalling that $g(y) \ne 0$) $f$ is differentiable at $0$, and then $f$ is differentiable everywhere, with $f'(y) = g(y) f'(0)$. Since $f'$ is not everywhere $0$, we must have $f'(0) \ne 0$ and $f'(y) \ne 0$.
Now the equation becomes
$$ f(x + y) - f(y) = f(x) f'(y)/f'(0) $$
Let $F = f/f'(0)$. Then the equation becomes
$$F(x+y) - F(y) = F(x) F'(y)$$
Taking the derivative with respect to $x$,
$$F'(x+y) = F'(x) F'(y)$$
That says $F'$ is a homomorphism from the additive group $\mathbb R$ into the multiplicative group $\mathbb R_+$. The only such homomorphisms that are measurable are of the form $F'(x) = \exp(c x)$ for constant $c$. If $c = 0$ we get $F(x) = x$ and $g(y) = 1$. Otherwise, integrating, $F(x) = k + c^{-1} \exp(c x)$ for some constant $k$. Substituting this in to the original equation and rearranging, I get
$$ \left( \exp(cy)-g(y)\right) \exp(cx) = c k g(y) + \exp(cy) $$
Since the left side can't depend on $x$, we must have $g(y) = \exp(cy)$, and then $c k + 1 = 0$. Thus the "non-official" solutions are $$ \eqalign{f(x) &= \dfrac{f'(0)}{c} \left(\exp(cx) - 1\right) \cr g(y) &= \exp(cy)\cr c &\ne 0, f'(0) \ne 0}$$
As $ f $ is strictly monotonic, there is $ c \in \mathbb R $ with $ f ( c ) \ne 0 $. Let $ a = f ( c ) $ and $ b = g ( c ) $. Note that by swapping $ x $ and $ y $ in $$ f ( x + y ) = f ( x ) g ( y ) + f ( y ) \text , \tag 0 \label 0 $$ you can deduce $$ f ( x ) g ( y ) + f ( y ) = f ( y ) g ( x ) + f ( x ) \text , $$ which for $ y = c $ gives $$ g ( x ) = \frac { b - 1 } a f ( x ) + 1 \text . \tag 1 \label 1 $$ Multiplying \eqref{0} by $ \frac { b - 1 } a $ and using \eqref{1} you get $$ g ( x + y ) = g ( x ) g ( y ) \text . \tag 2 \label 2 $$ Putting $ x = y = 0 $ in \eqref{2}, you have $ g ( 0 ) \in \{ 0 , 1 \} $. If $ g ( 0 ) = 0 $, then setting $ y = 0 $ in \eqref{2} shows that $ g $ is constantly zero; but that can't happen, since then by \eqref{1} you can deduce that $ f $ is constant, and that contradicts with $ f $ being strictly monotonic. Thus, $ g ( 0 ) = 1 $. Letting $ y = - x $ in \eqref{2} you get $ g ( x ) g ( - x ) = 1 $, which in particular shows that $ g ( x ) \ne 0 $. As Substituting $ \frac x 2 $ for both $ x $ and $ y $ in \eqref{2} we get $ g ( x ) = g \left( \frac x 2 \right) ^ 2 \ge 0 $, we can conclude that $ g $ is strictly positive at all points. This lets us define $ h : \mathbb R \to \mathbb R $ with $ h ( x ) = \log g ( x ) $. Then \eqref{2} shows that $ h $ satisfies Cauchy's functional equation. Also, as $ f $ is strictly monotonic, it is locally bounded, and thus by \eqref{1} and the definition of $ h $, $ h $ is locally bounded. By a well-known fact, this implies that there must be a constant $ k \in \mathbb R $ such that $ h ( x ) = k x $ for all $ x \in \mathbb R $. This means that we must have $ g ( x ) = \exp ( k x ) $ for all $ x \in \mathbb R $, and thus every $ g $ for which there is a suitable $ f $ must be of this form. It remains to check that for any function $ g $ of this form there is a suitable $ f $. To see this, it's sufficient to let $ f ( x ) = \exp ( k x ) - 1 $ in case $ k \ne 1 $, and $ f ( x ) = x $ in case $ k = 1 $. In both cases, it's straightforward to verify that $ f $ is strictly monotonic and \eqref{0} is satisfied.