Functions on $ ( 0 , + \infty ) $ satisfying $ f \big( f ( x y ) + 2 x y \big) = 3 x f ( y ) + 3 y f ( x ) $ 
Find all functions $ f : ( 0 , + \infty ) \to ( 0 , + \infty ) $ such that
$$ f \big( f ( x y ) + 2 x y \big) = 3 x f ( y ) + 3 y f ( x ) $$
for all $ x , y \in ( 0 , + \infty ) $.

I guess $ f ( x ) = 4 x $, but don't know how to do more. Please, give me some advice, thanks.

You can show that the only function $ f : ( 0 , + \infty ) \to ( 0 , + \infty ) $ satisfying
$$ f \big( f ( x y ) + 2 x y \big) = 3 x f ( y ) + 3 y f ( x ) \tag 0 \label 0 $$
for all $ x , y \in ( 0 , + \infty ) $ is $ f ( x ) = 4 x $. It's easy to see that this function is a solution. To see the converse, first note that letting $ a = f ( 1 ) $ and setting $ y = 1 $ in \eqref{0} we have
$$ f \big( f ( x ) + 2 x \big) = 3 a x + 3 f ( x ) \text . \tag 1 \label 1 $$
Substituting $ \frac x y $ for $ x $ in \eqref{0} we get
$$ f \big( f ( x ) + 2 x \big) = \frac { 3 x f ( y ) } y + 3 y f \left( \frac x y \right) \text , $$
which together with \eqref{1} gives
$$ a x + f ( x ) = \frac { x f ( y ) } y + y f \left( \frac x y \right) \text . \tag 2 \label 2 $$
Substituting $ \frac 1 y $ for $ y $ in \eqref{2} we get
$$ f ( x y ) = a x y + y f ( x ) - x y ^ 2 f \left( \frac 1 y \right) \text , $$
and as putting $ x = 1 $ in \eqref{2} we have
$$ y ^ 2 f \left( \frac 1 y \right) = 2 a y - f ( y ) \text , $$
we conclude that
$$ f ( x y ) = - a x y + x f ( y ) + y f ( x ) \text . \tag 3 \label 3 $$
Now, if we define $ g : \mathbb R \to \mathbb R $ with $ g ( x ) = \exp ( - x ) f ( \exp x ) - a $, then \eqref{3} can be simplified to
$$ g ( x + y ) = g ( x ) + g ( y ) $$
for all $ x , y \in \mathbb R $. Note that since $ f $ only takes positive values, by definition of $ g $ we have $ g ( x ) > - a $ for all $ x \in \mathbb R $. It's a well-known fact that the only additive function on $ \mathbb R $ which is bounded below on the whole of its domain, is the constant zero function. This means that $ f ( x ) = a x $ for all $ x \in ( 0 , + \infty ) $. As by letting $ x = 1 $ in \eqref{1} we have $ f ( a + 2 ) = 6 a $ and $ a $ is positive, we get $ a = 4 $, which completes the proof.
