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.

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 $a \geq 0$ .
