Find all functions $f:$ $\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ $ x^{2}(f(x)+f(y))=(x+y) f(f(x) y) $ Question -
(Czeck-Slovak, 2004)Find all functions $f:$ $\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$$
x^{2}(f(x)+f(y))=(x+y) f(f(x) y)
$$
for all $x, y >0 $
My work - 
by putting $y=x$ we get $f(xf(x))=xf(x)$ so $xf(x)$ is fixed point.
now by putting $x=1$ here we get $f(f(1))=f(1)$ and 
then putting $x=f(1)$ and $y=1$ we easily get that $f(1)=1$ ...
now my idea is that if we somehow proved that f is strictly decreasing then it should have only one fixed point so hence $xf(x)=1$ for all x..
but i am not able to show that f is indeed strictly decreasing ...can someone show it ???
and also i know by putting $x=1$ and using the fact that $f(1)=1$ we get our solution as $f(x)=1/x$ ..
but i want to solve this by proving that f is strictly decreasing...so can someone show it ..
thankyou
 A: Suppose there exists $2$ fixed points, $x$ and $y$. Then $x + y > 0$, so we get
$$
x^2(x + y) = (x + y)f(xy)
$$
and therefore $f(xy) = x^2$. Swapping $x$ and $y$, we get $f(xy) = y^2$. Therefore $x^2 = y^2$, and $x = y$. 
So, there can be at most $1$ fixed point. Now, you have shown that $1$ is a fixed point. But $xf(x)$ is a fixed point for all $x > 0$, therefore
$$
f(x) = \frac{1}{x}
$$
for all $x \in \mathbb R^+$. We must then check this solution. For all $x,y$ we have
$$
\begin{aligned}
x^2(f(x) + f(y)) &= x^2\left(\frac 1 x + \frac 1 y\right)\\
&= x + \frac{x^2}{y}\\
&= \frac{x}{y}\left(y + x\right)\\
&= (x + y)f\left(\frac{y}{x}\right)\\
&= (x + y)f(f(x)y)
\end{aligned}
$$
So $f\colon x \mapsto \frac{1}{x}$ is a solution, and is the only solution.
It is usually pretty hard to prove monotonicity in functional equations, without outright finding the function. To reduce the possible number of fixed points, it is however pretty common to plug in two fixed points into the equation, which makes the contibutions of $f$ "vanish".
