We are given the functional equation $f(x + y g(x)) = g(x) + x f(y)$ for all $x, y \in \mathbb{R}$. We aim to find all functions $f$ and $g$ from $\mathbb{R} \to \mathbb{R}$ that satisfy this equation.

First, assume $g(0) = 0$.

### Fact 1: $f(x) = g(x)$ for every $x$.
Proof:
Set $y = 0$ in the given equation:
$$f(x + 0 \cdot g(x)) = g(x) + x f(0) \implies f(x) = g(x) + x f(0).$$
Rewriting, we get:
$$g(x) = f(x) - x f(0).$$
In particular, if $x = 0$, then $g(0) = f(0)$. Since $g(0) = 0$, it follows that $f(0) = 0$. Therefore,
$$f(x) = g(x).$$

### Fact 2: Either $z = 0$ is the only zero of $f(z) = g(z) = 0$, or $f(x) = g(x) = 0$ for every $x$.
Proof:
Assume $f(z) = g(z) = 0$ for some $z$. Then, setting $x = z$ in the original equation, we get:
$$f(z + y g(z)) = g(z) + z f(y) \implies f(z) = 0 = z f(y).$$
Since this must hold for every $y$, if $z \neq 0$, then $f(y) = 0$ for all $y$. Hence, $f(x) = g(x) = 0$ for every $x$.

Thus, the solutions to the functional equation are:
1. $f(x) = g(x) = 0$ for all $x \in \mathbb{R}$.
2. $f(x) = g(x)$ for all $x \in \mathbb{R}$, where $f$ and $g$ are arbitrary functions satisfying $f(0) = 0$.

The answer is: $\boxed{f(x)}$ = g(x) = 0 $\text{}$ for all } x $\in \mathbb{R} \text{}$ or } f(x) = g(x) $\text{}$ with } f(0) = 0}.