Find all $f:\mathbb R\to\mathbb R$ such that $\forall x,y\in\mathbb R$ the given equality holds: $xf(y)+yf(x)=(x+y)f(x)f(y)$. 
Find all $f:\mathbb R\to \mathbb R$ such that $\forall x,y\in\mathbb R$ the given equality holds: $$xf(y)+yf(x)=(x+y)f(x)f(y)$$


My try:
whenever $y=0$, we have $$x\cdot f(0)\cdot(1-f(x))=0$$
So the solutions can be either generated by: $$f(x)=\begin{cases}c & c\in\mathbb R & x=0\\ 1 & x\neq 0\end{cases}$$
where $c$ generate an infinite amount of solutions, or it can be any function satisfying $f(0)=0$.
However, I can't seem to find a way to check such a solution on the initial equation (we have to check it since we've simply eliminated the impossible solutions and we've only found that the solutions can be somehow generated using the way I've shown).
Note that this generates the trivial solutions $f(x)=0$ and $f(x)=1$ as well.

Is my reasoning correct?
If so, how can I check whether all the generated functions satisfy the initial equation?

Off-topic, but do $f^2(x)$ and $f(x)^2$ denote the same concept? My teacher uses the former while my book uses the latter (for squaring $f(x)$). Which one is correct, or are both of them correct?
 A: You're most of the way there; now just show that your piecewise $f$ works for arbitrary $x$ and $y$.  You may find it easiest to break the analysis into four cases: $x,y\neq 0$; $x=0, y\neq 0$; $x\neq 0, y=0$; and $x, y=0$.  For instance, for the last case, you have $xf(y)+yf(x) = 0\cdot c+0\cdot c = 0 = (0+0)\cdot c\cdot c = (x+y)\cdot f(x)\cdot f(y)$.  You should find similar results in the other cases.
But note that not every function satisfying $f(0)=0$ satisfies the equation; what happens if you take e.g. $f(x) = \sin x$, $x=y=\frac\pi4$?  What you've shown is that every function with $f(0)=0$ satisfies the equation when we choose one of $x,y$ to be $0$, but the equation has to hold for arbitrary $x$ and $y$.  You may find it useful to set $x=y$ and see what the resultant restriction on $f(x)$ is.
Incidentally, the approach that you're taking is the typical way of solving these questions: find special values of $x$ or $y$ that simplify the equation considerably and turn it into a one-parameter equation.  For instance, taking $x=0$ here (obviously) simplifies the equation considerably; so does taking $x=y$.
A: Your solution is not complete. You have only identified a necessary condition. It is not obvious, not true, that they are sufficient conditions. 
