A functional equation solved with derivatives

Step 1. Taking $x=y=0$: $f(0)=2f(0)$, so $f(0)=0$.

Step 2: differentiating in $x$. For fixed $y$, $f'(x+y)=f'(x)+y$. With $x=0$: $f'(y)=f'(0)+y$. Let $a=f'(0)$.

Step 3: integrating. $f'(y)=a+y$ means $f(y)=ay+\frac{y^2}2+C$, and $f(0)=0$ gives $C=0$. So $f(x)=\frac{x^2}2+ax$.

Step 4: verification.
$$f(x+y)=\frac{(x+y)^2}2+a(x+y)=\frac{x^2}2+xy+\frac{y^2}2+ax+ay=f(x)+f(y)+xy.$$

Answer. The solutions are exactly the functions $f(x)=\dfrac{x^2}2+ax$, $a\in\mathbb{R}$.

Remark. Another route: $g(x)=f(x)-\frac{x^2}2$ satisfies Cauchy's equation $g(x+y)=g(x)+g(y)$, whose differentiable solutions are the linear functions.