$P(x,0)$ $ff(x)=f(x)(f(0)+1)$ Compare with original equation : $f(x+y)f(0)+xy=f(x)f(y)$ $P(0,0)$ $ff(0)=f(0)^2+f(0)$ $P(1,-1)$ $f(1)f(-1)=f(0)^2-x^2$ $P(f(0),-f(0))$ $\implies$ $f(0)(f(0+1)f(-f(0))=0$ $\text{First Case}$ $f(0)=0$ $f(x)f(y)=xy$ it is obvious that $f(1) \ne 0$ .Thus, $P(x,1)$ $\implies$ $f(x)=\frac{x}{c}$ and $c=1$ $\text{Second Case}$ $f(0)=-1$ from first equation $ff(x)=0$ for all real $x$. So, $f(x)f(y)+f(x+y)=xy$ $P(f(x),-f(x))$ $\implies$ $1=f(x)^2$ which is contradiction beacuse $f$ can't be a constant. $\text{Third Case}$ $f(-f(0))=0$ $P(-f(0),0$ $\implies$ $f(0)=0$ which leads to first case.

Likely the same as above. Denoting the given assertion by $P(x,y)$, $P(x,0)$ gives $f(f(x)) = f(x)(1+f(0))$ and taking $x=0$ here we get $f(f(0)) = f(0)+f(0)^2$. Next, $P(x,-x)$ gives $f(f(0)) = f(0) + f(x)f(-x)+x^2$, and taking $x=f(0)$ and recalling $f(f(0)) = f(0)+f(0)^2$ we obtain that \[ f(f(0))f(-f(0)) = f(0)(1+f(0))f(-f(0))=0. \]Thus, there are three cases to consider: (a) $f(0)=0$, (b) $f(0)=-1$, and (c) $f(-f(0)) = 0$. Suppose $f(0)=0$. Then, $f(f(x)) = f(x)(1+f(0))$, giving that $f(f(x))=f(x)$. Inserting this into the original equation, we get $f(x)f(y)=xy$ for all $x,y$. Taking $y$ to be an arbitrary non-zero constant, we find there is a $c$ for which $f(x)=cx$ for all $x$ (note that $x=0$ also satisfies this), which upon plugging gives $c=1$. Next, suppose $f(0)=-1$. Then, $f(f(x)) = 0$ for all $x$, yielding $f(x+y)+f(x)f(y)-xy=0$. Taking $f(x)$ in place of $x$ and $-f(x)$ in place of $y$, we find $-1+f(x)^2=0$, so $f(x)=\pm 1$ for all $x$, contradicting with $f(f(x))=0$ for all $x$. Lastly, we assume $f(-f(0))=0$. Using $f(f(x)) = f(x)(1+f(0))$ with $x=-f(0)$, we find that $f(0)=0$, going back to the first case.

Setting $y=0$ gives $f(f(x))=f(x)+f(x)f(0)$. So $f(x+y)+f(x+y)f(0)=f(x+y)+f(x)f(y)-xy$, thus $f(x+y)f(0)=f(x)f(y)-xy$. Take $x=f(0), y=-f(0)$. Then we get $f(0)^2=f(f(0))f(-f(0))+f(0)^2$, thus either $f(f(0))$ or $f(-f(0))$ is zero. Call it $\lambda$ Then $f(0)=f(f(\lambda))=0+0=0$, thus $f(0)=0$. $f(x)f(y)=xy$. Taking $x=y$ we get $f(x)=x$ or $f(x)=-x$. If both cases occur($f(a)=a, f(b)=-b, a,b \neq 0$), then $ab=-ab$ - a contradiction. So either $f(x) \equiv x$ or $f(x) \equiv -x$. By substituting them into the original equation, we get that only $f(x)=x$ works.

Let $P(x,y)$ denote the assertion Then $P(0,0)$ and $P(f(0),-f(0))$ gives that $f(f(0))f(-f(0))=0$, so there exist $c$ such that $f(c)=0$. $P(0,c)$ gives $f(0)=0$ and $P(x,0)$ gives $f(f(x))=f(x)$ and $P(x-1,0)$ gives $f(x-1)f(1)=x-1$ so $f(x)=\frac {x}{a}$ and $a=1$. Therefore the solution is $f(x)=x$ for all $x \in \mathbb{R}$.

https://artofproblemsolving.com/community/q1h3283769p30250342

The answer is $\boxed{f(x)=x}$, which can be verified to work. Claim: $0$ is in the range of $f$ Substitute $y=-x$ yielding $f(x)f(-x)=-x^2+f(f(0))-f(0)$, Let $d=f(f(0))-f(0)$. By letting $x=0$ we get that $0\leq d$. The claim follows by considering $x=\sqrt{d}$. Claim: $f(0)=0$ Consider an $x$ such that $f(x)=0$. Then letting $y=0$ gives $f(0)=0$. Claim: $f(f(x))=f(x)$ Just consider $y=0$. Claim: $f(x)f(y)=xy$ This follows from $f(f(x+y))=f(x+y)$. Now this is a standard equation and the result is all functions of the form $f(x)=cx$. Plugging into the previous claim yields $c=1$.

Solved with CyclicISLscelesTrapezoid, DottedCaculator The answer is $f(x)\equiv x$ only, which clearly works. Let $P(x,y)$ denote the assertion in the problem statement. $P(x,0)$ implies $f(f(x)) = f(x) + f(x)f(0)$, so subbing back into $P(x,y)$ yields $f(0)f(x+y) = f(x)f(y) - xy$. Call this $Q(x,y)$. We take two cases. Case 1: $f(0)=0$. Then $Q(x,y)$ reduces to $f(x)f(y)=xy$. Setting $x=y=1$ gives $f(1)=\pm 1$, and then setting $y=1$ gives $f(x)\equiv \pm x$. Note that $f(x)\equiv -x$ is not a solution. Case 2: $f(0)\neq 0$. Then $Q(f(0),-f(0))$ gives $f(f(0))f(-f(0))=0$, so $f$ has some root $c$. Then $Q(x-c,c)$ gives $f(0)f(x) = -c(x-c)$, so $f$ is linear, and we can check that the only linear solution is $f(x)\equiv x$.