VMO

By letting $y=0$ in (1), we have $f(x)\cdot f(0)=f(-1)+xf(0)$. We distinguish two cases regarding the value of $f(0)$.

Case 1. $f(0)\ne 0$ implies $f(x)=x+c$ for all $x\in \mathbb{R}$ with $c$ is constant, which is not a solution.

Case 2. $f(0)=0$ implies $f(-1)=0$. By plugging $x=y=1$ into (1), we get $f(1{)}^2=2f(1)$, so $f(1)=0$ or $f(1)=2$. From (1), the substitution $y=-1$ yields $$f(-y-1)=f(y),\forall y\in \mathbb{R}.(2)$$ Replacing $y$ by $-y-1$ in (1), we get $$f(x)f(-y-1)=f(-x(y+1)-1)+xf(-y-1)-(y+1)f(x).$$ Thus, it follows from (2) $$f(xy-1)+yf(x)=f(xy+x)-(y+1)f(x),\forall x,y\in \mathbb{R}.(3)$$ Let $x\ne -1$, replacing $x$ by $x+1$ and $y$ by $\frac{1}{x+1}$ in (3), we have $$f(x-1)=\frac{x-1}{x+1}f(x),\forall x\ne -1.$$ On the other hand, we set $y=1$ in (1) and get $$\begin{array}{rl}f(x)\cdot f(1)& =f(x-1)+x\cdot f(1)+f(x)\\ & =\frac{x-1}{x+1}\cdot f(x)+x\cdot f(1)+f(x),\forall x\ne -1.\end{array}(4)$$ If $f(1)=0$ then $f=0$ because (4) and $f(-1)=0$. If $f(1)=2$ then $f(x)=x(x+1)$ because (4) and $f(-1)=0$.

It is easy to see that $f(x)\equiv 0$ and $f(x)=x(x+1)$ satisfied (1). $\square$