Solve the functional equation $f(xf(y)+x)=xy+f(x)$

Find all $f:\mathbb{R}\rightarrow\mathbb{R}$ so that $f(xf(y)+x)=xy+f(x)$.

If you put $x=1$ it's easy to prove that f is injective.
Now putting $y=0$ you can get that $f(0)=0$.

$y=\frac{-f(x)}{x}$ gives us $f(\frac{-f(x)}{x})=-1$

Answer

Set $x = 1$ and $y = t - f(1)$,

then $f(1+f(t - f(1))) = t$, for all $t \in \mathbb{R}$, hence $f$ is a surjective function.

So, $\exists\, y_1 \in \mathbb{R}$, s.t., $f(y_1) = 0$

Then,

$$f(x) = f(x+f(y_1)x) = xy_1+f(x) \implies xy_1 = 0, \forall x \in \mathbb{R} \implies y_1 = 0$$

and $\exists\, y_2 \in \mathbb{R}$, s.t., $f(y_2) = -1$

Then,

$$0 = f(0) = f(x+xf(y_2)) = xy_2 + f(x)$$

i.e., $f(x) = -y_2x$ for $x \in \mathbb{R}$.

Plugging it back in the functional equation we may verify $y_2 = \pm 1$.