, 2011

Setting $y = x$, we obtain $2x f(xf(x)) = 2x^2 f(x)$, and so $f(xf(x)) = x f(x)$ since $x > 0$.

Now suppose $f(x) = f(y)$. Then
$$x^2(f(x) + f(y)) = f(yf(x))(x + y) = f(yf(y))(x + y) = y f(y)(x + y)$$
and so
$$2x^2 f(x) = (x y + y^2) f(y) \implies 2x^2 - x y - y^2 = (2x + y)(x - y) = 0 \implies x = y,$$
since all the quantities in the equations are positive and so $y \ne -2x$. So $f$ is injective. Moreover, setting $x = y = 1$, we obtain $f(f(1)) = f(1)$, and so injectivity implies that $f(1) = 1$.

Now set $x = 1$ in the given equation:
$$f(y)(y + 1) = 1 + f(y) \implies f(y) = \frac{1}{y},$$
and it is a simple matter to check that the solution $f(x) = \frac{1}{x}$ satisfies the given equation.