The functional equation $\big(1 + yf(x)\big)\big(1 - yf(x + y)\big) = 1$ for $f:\mathbb R^+\to\mathbb R^+$ Functional equation from USAMO 2010 preparation session:

Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that
$\big(1 + yf(x)\big)\big(1 - yf(x + y)\big) = 1$
for all $x, y \in \mathbb R^+$, where $\mathbb R^+$ is a set of all positive real numbers.

Well I don't really see what we can do. I mean I could have plugged in some numbers but the best one ($0$) which could have caused simplifications is out of reach and so are the negative numbers. Maybe if we could take two cases:

*

*Both factors are equal to $\pm 1$.

*The factors are reciprocals of each other.

In the first case $+1$ in fact is only possible when $f(x)=0$ $ \forall x$, which isn't possible as $0$ isnt in the codomain. For $-1$ we would get $f(x+y)=-f(x)$, contradiction again.
So we must have the two factors being reciprocals. Well now what? I'm stuck here. Clearly my approach is not just unprofessional, it's bad too. How can this thing be solved?
EDIT:
After considering @Yesit'sme's comment, I retried the problem and would like to present a solution. PLEASE DO TELL ME WHETHER IT IS CORRECT OR NOT. Here we go.
Since $x,y \in \mathbb R^+$, we may without restriction assume $x,y \neq 0$.
Now from given,
$\begin{align}
\big(1 + yf(x)\big)\big(1 − yf(x + y)\big) &= 1 \\
1 − yf(x + y)&= \frac{1}{1 + yf(x)} \\
1-\frac{1}{1 + yf(x)} &= yf(x + y) \\ 
\frac{1+yf(x)-1}{1 + yf(x)} &= yf(x + y) \\
\frac{f(x)}{1+yf(x)} &= f(x+y)= \frac{f(y)}{1+xf(y)} \tag 1 \label 1
\end{align}$
$\forall x,y \in \mathbb R^+$.
The last equation follow from symmetry (or plugging in $y+x$ into $f$).
Now by plugging in $y=1$ we see that,
$ f(x+1) =\frac{f(x)}{1+f(x)}<f(x), \forall x \in \mathbb R^+$.
The last inequality follows from the fact that,
$\begin{align} f(x)+1 &>1 \\ 
1 &>\frac{1}{f(x)+1} \\ 
f(x)&>\frac{f(x)}{f(x)+1}=f(x+1) \end{align}$
As $f(x) \in \mathbb R^+$.
Thus $f(x)$ is in fact decreasing. We now define a new function $Q$ such that,
$f(x)=\frac{1}{Q(x)}$
where $Q$ is strictly increasing $\forall x$.
Plugging this into \eqref{1} we get,
$\begin{align} \frac{\frac{1}{Q(x)}}{1+\frac{y}{Q(x)}} &= \frac{\frac{1}{Q(y)}}{1+\frac{x}{Q(y)}} \\
\frac{1}{Q(x)+y} &= \frac{1}{Q(y)+x} \\ 
Q(x)+y &=Q(y)+x \\
Q(y)-y &=Q(x)-x=k \\ \end{align} $
Where $k \in \mathbb R$ is a constant. This gives,
$\begin{align}\frac{1}{f(x)} &= x+k \\
\therefore f(x) &= \frac{1}{x+k} \blacksquare. \\ \end{align} $
Plugging this into the original equation we see that the equation is satisfied and hence the solution is complete.
(I didn't show the checking part as typing out this much already took a boatload of time. Hope you understand. It does satisfy though, I have checked it by hand.)

Since $x,y \in \mathbb R^+$, we may without restriction assume $x,y \neq 0$.
Now from given,
$ \begin{align} (1 + yf(x))(1 − yf(x + y)) &= 1 \\
1 − yf(x + y)&= \frac{1}{1 + yf(x)} \\
1-\frac{1}{1 + yf(x)} &= yf(x + y) \\ 
\frac{1+yf(x)-1}{1 + yf(x)} &= yf(x + y) \\
\frac{f(x)}{1+yf(x)} &= f(x+y)= \frac{f(y)}{1+xf(y)} \tag 1 \label {eqn1} \\ \end{align} $
$\forall x,y \in \mathbb R^+$.
The last equation follow from symmetry (or plugging in $y+x$ into $f$).
Since,
$f ( x + y ) = \frac { f ( x ) } { f ( x ) + y } = \frac 1 { \frac 1 { f ( x ) } + y }<\frac{1}{\frac{1}{f(x)}}=f(x)$
The inequality follows from the fact that,
$\begin{align} y &>0 \\ 
\frac{1}{f(x)}+y &>\frac{1}{f(x)} \\ 
\frac{1}{\frac{1}{f(x)}} &> \frac{1}{\frac{1}{f(x)}+y} \end{align}$
As $f(x) \in \mathbb R^+$.
Thus, $f(x)$ is in fact decreasing.
We now define a new function $Q$ such that,
$f(x)=\frac{1}{Q(x)}$
Since $f \neq 0$.
Plugging this into \eqref{eqn1} we get,
$\begin{align} \frac{\frac{1}{Q(x)}}{1+\frac{y}{Q(x)}} &= \frac{\frac{1}{Q(y)}}{1+\frac{x}{Q(y)}} \\
\frac{1}{Q(x)+y} &= \frac{1}{Q(y)+x} \\ 
Q(x)+y &=Q(y)+x \\
Q(y)-y &=Q(x)-x=k \\ \end{align} $
Where $k \in \mathbb R$ is a constant. This gives,
$\begin{align}\frac{1}{f(x)} &= x+k \\
\therefore f(x) &= \frac{1}{x+k} \blacksquare. \\ \end{align} $
Plugging this into the original equation we see that the equation is satisfied and hence the solution is complete.
