Problem: Solve over positive real numbers the functional equation
\[ f(x) = f(f(f(x)) + y) + f(xf(y)) f(x+y). \]

Solution: 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 \texttt{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
\[ Q(a,b) : \qquad f(a) \ge f(b) \implies f(f(b)) \ge a. \]
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:
\begin{claim*}
  Any function $f \colon \RR_{>0} \to \RR_{>0}$ obeying
  statement $Q$ satisfies $f^2(x) = f^4(x)$.
\end{claim*}
\begin{proof}
  From $Q(t,t)$ we get
  \[ f^2(t) \ge t \qquad \text{ for all } t > 0. \]
  So this already implies $f^4(x) \ge f^2(x)$ by choosing $t = f^2(x)$.
  It also gives $f(x) \le f^3(x) \le f^5(x)$
  by choosing $t = f(x)$, $t = f^3(x)$.

  Then $Q(f^4(x), x)$ is valid and gives $f^2(x) \ge f^4(x)$, as needed.
\end{proof}

\begin{claim*}
  The function $f$ is injective.
\end{claim*}
\begin{proof}
  Suppose $f(u) = f(v)$ for some $u > v$.
  From $Q(u,v)$ and $Q(v,u)$ we have $f^2(v) \ge u$ and $f^2(u) \ge v$.
  Note that for all $x > 0$ we have statements
  \begin{align*}
    P(f^2(x), u) &\implies f^3(x) = f(x+u) + f(xf(u)) f(x+u) = (1+f(xf(u)))f(x+u) \\
    P(f^2(x), v) &\implies f^3(x) = f(x+v) + f(xf(v)) f(x+v) = (1+f(xf(v)))f(x+v).
  \end{align*}
  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.
\end{proof}

Since $f$ is injective, we obtain that $f^2(x) = x$.
Thus $P(x,y)$ now becomes the statement
\[ P(x,y) : \qquad f(x) = f(x+y) \cdot \Big[ 1+f(xf(y)) \Big]. \]
In particular
\[ P(1,y) \implies f(1+y) = \frac{f(1)}{1+y} \]
so $f$ is determined on inputs greater than $1$.
Finally, if $a,b > 1$ we get
\[ P(a,b) \implies \frac 1a = \frac{1}{a+b} \cdot
  \left[ 1 + f\left( \frac ab f(1) \right) \right] \]
which is enough to determine $f$ on all inputs, by varying $(a,b)$.