Determining all functions satisfying the functional equation $ f \big( x + y f ( x ) \big) + f ( x y ) = f ( x ) + f ( 2019 y ) $ I came up with the following functional equations problem, which is from Romanian Master of Mathematics 2019 Day 2, Problem 5 (still lack of solution by 2019 February 28th). I found this problem interesting, but I do not know how to do it. I want to know in general, how can one deal with such a functional problem. Are there any recommend books, lecture notes and etc.

Determine all functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \big( x + y f ( x ) \big) + f ( x y ) = f ( x ) + f ( 2019 y ) $$
for all real numbers $ x $ and $ y $.

Through some investigation, I guess, $ f ( x ) = \text {some constant}$ or $ f ( x ) = 2019 - x $ assuming it is a linear function. I do not know whether those are the only solutions.
 A: Consider $ N \in \mathbb R \setminus \{ 0 \} $ (in the case of the above problem, $ N = 2019 $). There are three types of functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \big( x + y f ( x ) \big) + f ( x y ) = f ( x ) + f ( N y ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $:
$$ f ( x ) = N - x \text ; \tag i \label i $$
$$ f ( x ) = c \text { for an arbitrary constant } c \text ; \tag {ii} \label {ii} $$
$$ f ( x ) = 0 \text { for } x \ne 0 \text {, and } f ( 0 ) \text { is arbitrary.} \tag {iii} \label {iii}$$
A straightforward check shows that all three types satisfy the equation, hence we just need to show that they are the only ones.
First of all, substituting $ N x $ for $ x $ in \eqref{0}, we arrive at the equation
$$ f \big( N x + y f ( N x ) \big) + f ( N x y ) = f ( N x ) + f ( N y ) \text . $$
After a change $ g ( x ) = \frac { f ( N x ) } N $ this eqation reads
$$ g \big( x + y g ( x ) \big) + g ( x y ) = g ( x ) + g ( y ) \tag 1 \label 1 $$
for all $ x , y \in \mathbb R $, which does not depend on $ N $. Now we investigate the corresponding functions $ g $.
Setting $ x = 1 $ in \eqref{1} we get $ g ( 1 + y g ( 1 ) ) = g ( 1 ) $. If $ g ( 1 ) \ne 0 $, then $ 1 + y g ( 1 ) $ attains all real values, so we arrive at the answer \eqref{ii}. Otherwise, $ g ( 1 ) = 0 $, and by setting $ y = 1 $ in \eqref{1} we get $ g \big( x + g ( x ) \big) = 0 $. If $ a = 1 $ is the unique real number with $ g ( a ) = 0 $, then we obtain $ x + g ( x ) = 1 $, whence $ g ( x ) = 1 - x $, which falls into \eqref{i}. Also, if $ g ( x ) = 0 $ for all $ x \ne 0 $, we get the remaining answer \eqref{iii}. Hence in the sequel we assume that $ g ( 1 ) = 0 $, $ g ( a ) = 0 $ for some $ a \ne 1 $, and also there exists $ s \ne 0 $ with $ g ( s ) \ne 0 $, and arrive at a contradiction.
If $ b $ is an arbitrary zero of $ g $, then by plugging $ x = b $ in \eqref{1} we get $ g ( b y ) = g ( y ) $. Recalling that $ g ( g ( 0 ) ) = g \big( 0 + g ( 0 ) \big) = 0 $, we obtain
$$ g \big( g ( 0 ) y \big) = g ( a y ) = g ( y ) \text . \tag 2 \label 2 $$
Now, we show that $ g $ is $ p $-periodic, where $ p = ( a - 1 ) s $. Indeed, setting $ x = a s $ in \eqref{1} and using \eqref{2}, we get
$$ g \big( a s + y g ( s ) \big) = g \big( a s + y g ( a s ) \big) = g ( a s ) + g ( y ) - g ( a s y ) = g ( s ) + g ( y ) - g ( s y ) = g \big( s + y g ( s ) \big) \text , $$
which proves the required periodicity, since $ y g ( s ) $ attains all real values. Letting $ x = p $ in \eqref{1} and using periodicity we get $ g \big( y g ( 0 ) \big) + g ( p y ) = g ( 0 ) + g ( y ) $. By \eqref{2}, we arrive at $ g ( p y ) = g ( 0 ) $, which leads to the contradiction $ 0 = g ( 1 ) = g \big( p \cdot p ^ { - 1 } \big) = g ( 0 ) = g \big( p \cdot \frac s p \big) = g ( s ) \ne 0 $.
