Continuous Functions Problems I'm not sure how to solve continuous functions problems. So, could someone please explain these three problems with a solution?
Find all continuous functions $f$ for $x>0$ such that $f(xy)=xf(y)+yf(x)$.
Find all continuous functions $f$ for $x>0$ such that $f(x+y)=\frac{f(x)f(y)}{f(x)+f(y)}$.
Find all continuous real-valued functions $f$ satisfying $f(x+y)=f(x)+f(y)+f(xy)$
Are there any special function properties that maybe I don't know to solve these? Thanks!
 A: Here is a hint for the first one. First, I'll give a "trick," then explain where it came from.
For any solution $f\colon(0,\infty)\to\mathbb{R}$ to the first functional equation, define $g\colon \mathbb{R}\to \mathbb{R}$ by
$$
g(x) \stackrel{\rm def}{=} e^{-x} f(e^x).
$$
Then $g$ is continuous on $\mathbb{R}$, given $g$ we can easily get back $f$ since $f(x) = x g(\ln x)$ for $x>0$, and, for any $x,y\in\mathbb{R}$,
$$\begin{align}
g(x+y) &= e^{-(x+y)}f(e^{x+y})
= e^{-(x+y)}f(e^{x}e^y)
= e^{-(x+y)}\left(e^xf(e^y)+e^yf(e^x)\right) \\
&= e^{-y}f(e^{y})+e^{-x}f(e^{x})
= g(x)+g(y)
\end{align}$$
Therefore, it is sufficient to solve the following:

Find all continuous functions $g\colon\mathbb{R}\to\mathbb{R}$ such that
  $$
g(x+y)=g(x)+g(y), \qquad \forall x,y\in\mathbb{R}.
$$

Hopefully, you should have a much easier time to do this — this is a standard exercise.

Now, where did that come from? First, $f$ was defined on the positive reals only. This is a bit fishy: between this and the multiplicative aspects of the functional equation, it strongly hints that there should be some logarithms/exponentials involved. And indeed, setting $x=e^u$, $y=e^v$ for any $u,v\in\mathbb{R}$, we get
$$f(e^{u+v})=f(e^ue^v)=f(xy) = xf(y)+yf(x) = e^u  f(e^v) + e^v f(e^u)$$ for any $u,v$. This looks a bit nicer, but we'd like to "decouple" $u$ and $v$ in the equation. The simplest way to do so is to multiply both sides by $e^{-u} e^{-v}$, getting
$$e^{-(u+v)}f(e^{u+v}) = e^{-v}  f(e^v) + e^{-u} f(e^u)$$
which now looks much more manageable, and suggests to "set" $g(u) = e^{-u}f(u)$.
