Find all functions ${\rm f} :{ \mathbb R}_{+}\to{ \mathbb R}_{+}$ Find all functions ${\rm f}:{\mathbb R}_{+} \to {\mathbb R}_{+}$
, such that $\forall\ x,y \in \mathbb R_+$ the equation
$$
\big(1 + y{\rm f}\left(x\right)\big)\big(1 - y{\rm f}\left(x + y\right)\big) = 1
$$ is fulfilled.
 A: Assume $f$ is of the given form. Let $g(x)=\frac{1}{f(x)}$.
Plugging this into the equation gives
$$\left(1+\frac{y}{g(x)}\right)\left(1-\frac{y}{g(x+y)}\right)=1$$
Multiplying by $g(x)g(x+y)$, factoring out and subtracting $g(x)g(x+y)$ we obtain
$$yg(x+y)-y^2-yg(x)=0$$
Dividing by $y>0$ gives
$$g(x+y)=g(x)+y$$
This implies $\frac{g(x+y)-g(x)}{y}=1$ for all $x,y\in\mathbb{R}_+$.
In particular, $\lim_{y\rightarrow 0}\frac{g(x+y)-g(x)}{y}=1$ (*).
Hence $g$ is differentiable on $\mathbb{R}^+$ with derivative $1$. 
By the fundamental theorem of calculus, $g(x)=x+c$ for some $c\ge 0$, that is
$$f(x)=\frac{1}{x+c}$$ 
for $c\in\mathbb{R}_{\ge 0}$. These are also really solutions.
Important note (*):
We have the equation $g(x+y)-g(x)=y$ only for positive $y$. Therefore, the limit is a priori only a right-sided limit. This is a problem, because we can then only conclude right-differentiability, which is useless.
However, we can extend the equation to small negative $y$, thus making the limit two-sided.

Claim: Let $x>0$ be fixed. For all $y$ with $-x<y<0$ (that is: $y$ is negative and has small enough absolute value), we have
  $$\frac{g(x+y)-g(x)}{y}=1$$

Proof:
Plugging in $-y>0$ for $y$ and $x+y>0$ for $x$ into the original equation gives
$$1=\frac{g(x)-g(x+y)}{-y}=\frac{g(x+y)-g(x)}{y}$$ $\square$
A: Correct me if I'm very wrong.
$$(1+y f(x))(1-y f(x+y))=1 \implies \frac{f(x+y)-f(x)}{y}=-f(x)f(x+y) < 0$$
So $f$ is strictly decreasing.
If $f$ was not continous at some point $x_1 \in \mathbb{R}_+$, it would have to be a jump discontinuity downwards and from the restirction it must follow that there's a jump discontinuity at every positive $y$ greater than $x_1$. But $x_1=x+y$ for many $x,y \in \mathbb{R}_+$, and again it follow from the restriction that there's a jump discontinuity at $x$. given $x$ is arbitrary and smaller than $x_1$, it follows that there's a jump discontinuity at every positive $x$ smaller than $x_1$. Finally, we conclude $f$ is jump discontinous in all It's domain, and coupled with the fact $f$ is bounded above (strictly decreasing), and bellow, we have a contradiction (To be elaborated on later).
SO... $f$ is continous! And:
$$lim_{\Delta x \rightarrow 0^{+}}  \frac{f(x+\Delta x)-f(x)}{\Delta x} = lim_{\Delta x \rightarrow 0^{+}} -f(x)f(x+\Delta x)=-f(x)^2$$
Again, $f$ is continous, so for $x$ taken as a positive constant, $-f(x)f(x+\Delta x)$ is a continous function of $\Delta x$ for all $\Delta x > -x$. Hence:
$$lim_{\Delta x \rightarrow 0^{-}} \frac{f(x+\Delta x)-f(x)}{\Delta x} = lim_{\Delta x \rightarrow 0^{-}} -f(x)f(x+\Delta x) = lim_{\Delta x \rightarrow 0^{+}} -f(x)f(x+\Delta x) = lim_{\Delta x \rightarrow 0^{+}}  \frac{f(x+\Delta x)-f(x)}{\Delta x}$$
So $f$ is even differentiable, and:
$$f'(x)=-f(x)^2$$
And the only solutions are $f(x)=\frac{1}{x+a}$, for any constant $a \geq 0$ .
