Solve over positive real numbers the functional equation  
$$ f(x)=f(f(f(x))+y)+f(x f(y)) f(x+y) $$
The answer is $f(x) \equiv c / x$ for any $c>0$. This works, so we'll prove this is the only solution. The following is based on the solution posted by pad on AoPS. In what follows, $f^{n}$ as usual denotes $f$ iterated $n$ times, and $P(x, y)$ is the given statement. Also, we introduce the notation $Q$ for the statement 
$$ Q(a, b): \quad f(a) \geq f(b) \Longrightarrow f(f(b)) \geq a . $$
To see why this statement $Q$ is true, assume for contradiction that $a > f(f(b))$; then consider $P(b, a - f(f(b)))$ to get a contradiction. The main idea of the problem is the following: Claim - Any function $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ obeying statement $Q$ satisfies $f^{2}(x) = f^{4}(x)$. 
$$ f^{2}(t) \geq t \quad \text{for all } t > 0 $$
So this already implies $f^{4}(x) \geq f^{2}(x)$ by choosing $t = f^{2}(x)$. It also gives $f(x) \leq f^{3}(x) \leq f^{5}(x)$ by choosing $t = f(x), t = f^{3}(x)$. 
Then $Q(f^{4}(x), x)$ is valid and gives $f^{2}(x) \geq f^{4}(x)$, as needed. 
Claim - The function $f$ is injective. 
$$ \begin{aligned} & P(f^{2}(x), u) \Longrightarrow f^{3}(x) = f(x + u) + f(x f(u)) f(x + u) = (1 + f(x f(u))) f(x + u) \\ & P(f^{2}(x), v) \Longrightarrow f^{3}(x) = f(x + v) + f(x f(v)) f(x + v) = (1 + f(x f(v))) f(x + v) . \end{aligned} $$
It follows that $f(x + u) = f(x + v)$ for all $x > 0$. This means that $f$ is periodic with period $T = u - v > 0$. However, this is incompatible with $Q$, because we would have $Q(1 + nT, 1)$ for all positive integers $n$, which is obviously absurd. 
Since $f$ is injective, we obtain that $f^{2}(x) = x$. Thus $P(x, y)$ now becomes the statement 
$$ P(x, y): \quad f(x) = f(x + y) \cdot [1 + f(x f(y))] . $$
In particular 
$$ P(1, y) \Longrightarrow f(1 + y) = \frac{f(1)}{1 + y} $$
so $f$ is determined on inputs greater than 1. Finally, if $a, b > 1$ we get 
$$ P(a, b) \Longrightarrow \frac{1}{a} = \frac{1}{a + b} \cdot \left[1 + f\left(\frac{a}{b} f(1)\right)\right] $$
which is enough to determine $f$ on all inputs, by varying $(a, b)$.