Find all solutions to the functional equation $f(x+y)-f(y)=\frac{x}{y(x+y)}$ 
Find all solutions to the functional equation
  $f(x+y)-f(y)=\cfrac{x}{y(x+y)}$

I've tried the substitution technique but I didn't really get something useful.
For $y=1$ I have  
$F(x+1)-F(1)=\cfrac{x}{x+1}$
A pattern I've found in this example is that if I iterate $n$ times the function $g(x)=\cfrac{x}{x+1}$  I have that $g^n(x)=\cfrac{x}{nx +1}$ ,which may be a clue about the general behaviour of the function ( ?) .
I am really kinda of clueless ,it seems like the problem is calling some slick way of solving it.
Can you guys give me a hint ?
 A: $f(x+y)-f(y)
=\cfrac{x}{y(x+y)}
=\frac1{y}-\frac1{x+y}
$
so
$f(x+y)+\frac1{x+y}
=f(y)+\frac1{y}
$.
Therefore,
$f(x)+\frac1{x}$
is constant,
so
$f(x)
=d-\frac1{x}
$
for some $d$.
Substituting this,
the $d$s cancel out,
so any $d$ works,
and the solution is
$f(x)
=d-\frac1{x}
$
for any $d$.
A: Hint: Let $x=1-y$, hence $x+y=1$.
Then we get $$f(1)-f(y)=\frac{1-y}{y}$$
I will include the full solution in a spoiler, since you only asked for a hint.

  This gives $f(0)=c$ and $f(1)=d$, then $f(y)=\frac{y-1}{y}+d$ for all other $y$. This simplifies to $f(0)=c$ and $f(y)=\frac{y-1}{y}+d$ for all other $y$.  We have to check whether every such function satisfies. Note that neither $y$ nor $x+y$ can ever be zero, so $f(0)$ can indeed be anything.  Now we have to check the other values: Does  $$\left(\frac{x+y-1}{x+y} + d\right) - \left(\frac{y-1}{y} +d \right) = \frac{y}{y(x+y)} ?$$  $$\frac{y(x+y-1)-(y-1)(x+y)}{y(x+y)} = \frac{y}{y(x+y)} ?$$  It turns out that the functions do satisfy the functional equation. 

A: Another approach: I assume that $f$ is a function of a single real variable.Write the defining equation as $\frac{f(y+x) - f(y)}{(y+x)- y} = \frac{1}{y(x+y)}$
(for $x,y,(x+y) \neq 0).$ Take the limit as $x \to 0$. On the one hand this limit is $\frac{1}{y^{2}}.$ On the other hand, the limit is, by definition, the derivative $f^{\prime}(y)$, (and we have proved that this exists for $y \neq 0$).
An antiderivative of $\frac{1}{y^{2}}$ is $\frac{-1}{y}$. Hence $f(y) = \frac{-1}{y} + C$ for a constant $C$, as long as $y \neq 0$. (Strictly speaking this proves that any function $f$ which satisfies the equation has to be of the described form. It should be checked that such functions satisfy the equation- but they do).
