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

$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 $d$.

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.

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).