9. A3 (BLR) Find all pairs of functions \( f: \mathbb{R} \rightarrow \mathbb{R}, g: \mathbb{R} \rightarrow \mathbb{R} \) such that 
\[ f(x+g(y))=x f(y)-y f(x)+g(x) \quad \text{for all } x, y \in \mathbb{R} \]
9. Let us first solve the problem under the assumption that \( g(\alpha) = 0 \) for some \( \alpha \). Setting \( y = \alpha \) in the given equation yields \( g(x) = (\alpha + 1) f(x) - x f(\alpha) \). Then the given equation becomes \( f(x + g(y)) = (\alpha + 1 - y) f(x) + (f(y) - f(\alpha)) x \), so setting \( y = \alpha + 1 \) we get \( f(x + n) = m x \), where \( n = g(\alpha + 1) \) and \( m = f(\alpha + 1) - f(\alpha) \). Hence \( f \) is a linear function, and consequently \( g \) is also linear. If we now substitute \( f(x) = a x + b \) and \( g(x) = c x + d \) in the given equation and compare the coefficients, we easily find that 
\[ f(x) = \frac{c x - c^2}{1 + c} \quad \text{and} \quad g(x) = c x - c^2, \quad c \in \mathbb{R} \backslash \{-1\} \]
Now we prove the existence of \( \alpha \) such that \( g(\alpha) = 0 \). If \( f(0) = 0 \) then putting \( y = 0 \) in the given equation we obtain \( f(x + g(0)) = g(x) \), so we can take \( \alpha = -g(0) \). Now assume that \( f(0) = b \neq 0 \). By replacing \( x \) by \( g(x) \) in the given equation we obtain \( f(g(x) + g(y)) = g(x) f(y) - y f(g(x)) + g(g(x)) \) and, analogously, \( f(g(x) + g(y)) = g(y) f(x) - x f(g(y)) + g(g(y)) \). The given functional equation for \( x = 0 \) gives \( f(g(y)) = a - b y \), where \( a = g(0) \). In particular, \( g \) is injective and \( f \) is surjective, so there exists \( c \in \mathbb{R} \) such that \( f(c) = 0 \). Now the above two relations yield 
\[ g(x) f(y) - a y + g(g(x)) = g(y) f(x) - a x + g(g(y)) \]
Plugging \( y = c \) in (1) we get \( g(g(x)) = g(c) f(x) - a x + g(g(c)) + a c = k f(x) - a x + d \). Now (1) becomes \( g(x) f(y) + k f(x) = g(y) f(x) + k f(y) \). For \( y = 0 \) we have \( g(x) b + k f(x) = a f(x) + k b \), whence 
\[ g(x) = \frac{a - k}{b} f(x) + k. \]
Note that \( g(0) = a \neq k = g(c) \), since \( g \) is injective. From the surjectivity of \( f \) it follows that \( g \) is surjective as well, so it takes the value 0.