Find all real functions $f:\mathbb {R} \to \mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$

You proved that $f(0)=0$, we can continue from here

Suppose that there exist $k \neq 0$ such that $f(k)=0$. Then plugging $x=k,y=y-k$ gives, $f(k^2) = f(k^2 + (y-k)f(k)) = kf(k+(y-k))=kf(y)$, which means that $f(x)$ is a constant function and so $f(x)=0$.

Now suppose that there exist no $k \neq 0$ such that $f(k)=0$. Then plugging $x=x,y=-x$ gives $ f(x^2-xf(x)) = xf(0)= 0 $, which by assumption means $x^2=xf(x)$ or $f(x)=x$.

So we conclude that, possible functions are $f(x)=x,0 \forall x \in \mathbb{R} $

As you said $f(0)=0$ $$f(x^2+y.f(x))=x.f(x+y)$$ Take $y=0\implies f(x^2)=xf(x)$. So for $p>0$(p for positive,also I took this partition because $x^{1/2n}\mid n\in\mathbb N$ is defined only for $x>0$): $$f(p^2)=p(\sqrt pf(\sqrt p))=\lim_{n\to\infty}p^{\displaystyle \left(\sum_{k=0}^{n}\frac1{2^k}\right)}f\left(p^{1/n}\right)=p^2f(1)\\f(p)=kp\tag{$p>0$}$$ Now to find the function for $n<0$(n for negative): $$f((-x)^2)=f(x^2)\implies -xf(-x)=xf(x)\implies f(x)+f(-x)=0$$ $$f(n)=-f(-n)=-(-kn)=kn\tag{$n<0$}$$ So $f(x)$ is odd and we can say that it is: $$f(x)=kx$$