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
$$
Q(a, b): \quad f(a) \geq f(b) \Longrightarrow f(f(b)) \geq a \text {. }
$$

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 $f^{2}(x)=f^{4}(x)$.

Proof. From $Q(t, t)$ we get
$$
f^{2}(t) \geq t \quad \text { for all } t>0 \text {. }
$$

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 $Q\left(f^{4}(x), x\right)$ is valid and gives $f^{2}(x) \geq f^{4}(x)$, as needed.
Claim - The function $f$ is injective.

Proof. Suppose $f(u)=f(v)$ for some $u>v$. From $Q(u, v)$ and $Q(v, u)$ we have $f^{2}(v) \geq u$ and $f^{2}(u) \geq v$. Note that for all $x>0$ we have statements
$$
\begin{array}{l}
P\left(f^{2}(x), u\right) \Longrightarrow f^{3}(x)=f(x+u)+f(x f(u)) f(x+u)=(1+f(x f(u))) f(x+u) \\
P\left(f^{2}(x), v\right) \Longrightarrow f^{3}(x)=f(x+v)+f(x f(v)) f(x+v)=(1+f(x f(v))) f(x+v) .
\end{array}
$$

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 $Q(1+n T, 1)$ 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 $(a, b)$.