f(0)=3f(0)=>f(0)=0. y=-x=>f(-x²)=f(x)+f(-x)+f(-x²) for all real x so f(-x)=-f(x). Make y->-y to get f(x-y-xy)=f(x)-f(y)-f(xy). We add this to the initial relation and we get f(x+y+xy)+f(x-y-xy)=2f(x). For all a there obviously exists a y\in R s.t. y+xy=a, so we can rewrite it as f(x+y)+f(x-y)=2f(x) for all x and y\in R. If we make y=0 we get f(2x)=2f(x), so f(x+y)+f(x-y)=f(2x) so for all a and b\in R we have f(a)+f(b)=f(a+b). f is additive and continuous and it's easy to get from here that f(x)=f(1)*x for all reals x.

Can Jensen's be used to solve this in some way? Maybe something about $ f$ being convex?