abeker wrote:

Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
$f(x^2 + f(x)f(y)) = xf(x + y)$ for all real numbers $x$ and $y$ .

Assume that $f(x)=c$ for $c$ any constant.
$c=cx \; \forall x \in \mathbb R \implies c=0 \implies f(x) \equiv 0 \; \forall x \in \mathbb R$ Now take that $f$ is not constant.
Let $P(x,y)$ the assertion of the given FE.
$P(0,x)$
$f(f(x)f(0))=0 \; \forall x \in \mathbb R \implies f(0)=0$ $P(x,0)$
$f(x^2)=xf(x) \; \forall x \in \mathbb R$ $P(x,-x)$
$f(x^2+f(x)f(-x))=0 \; \forall x \in \mathbb R$ Now assume that exists more than 1 cero no the function.
Let $\mathbb C$ the set of the ceros of $f$ and take $c \in \mathbb C$ then $f(c)=0$ .
$P(x,c)$
$f(x^2)=xf(x+c) \; \forall x \in \mathbb R$ Then $f$ is periodic at $c$ .
Note that $c^2$ is also a cero by $f(x^2)=xf(x)$ .
Then:
$cf(2c)=0 \implies c=0 \; \text{or} \; f(2c)=0$ Now take $f(2c)=0$ then $f(4c^2)=0$
In fact if we skip the case when $c=0$ then $f(n \cdot c)=0 \; \forall n \in \mathbb Z$
$P(-x,0)$
$f(x^2)=xf(x)=-xf(-x) \implies f(-x)=-f(x)$ Assume that exists $a,b \ne n \cdot c$ such that $f(a)=f(b)$ then i will prove that $a=b$ .
$P(a,-b)+P(-b,a)$
$f(a^2-f(a)^2)+f(b^2-f(b)^2)=(a-b)f(a-b) \implies a=b$ Then $f$ is injective.
That means exists an only $0$ and hence $c=0$ (contradiction!!) (We assumed that exists more than 1 cero, thats why contradiction)
Hence $f(0)=0$ has an unique $0$
The same proof for get $f$ injective.
Then:
$x^2-f(x)^2=0 \implies f(x)= \pm x$ Then the solutions are:
$\boxed{f(x)=x \; \forall x \in \mathbb R}$

$\boxed{f(x)=-x \; \forall x \in \mathbb R}$

$\boxed{f(x) \equiv 0 \; \forall x \in \mathbb R}$

Thus we are done :blush: