What are the continuous functions $ x f(y)+y f(x)=(x+y) f(x) f(y) ? $ question - 

What are the continuous functions on $\mathbb{R}$ which are solutions of the equation
$$
x f(y)+y f(x)=(x+y) f(x) f(y) ?
$$

my try - 

by putting $y=x$ i get $f(x)=0$ or $1$ for all $x$ not equal to $0$...
now my answer is same as mention in book but i think it is incorrect because they are not valid for all $x$ ??? 

can someone tell how to fix this hole with the help of continuity ...
i know this is simple question but i want to clear my doubt...

thankyou
Wit $x=y$, we arrive at the necessary condition
$$2xf(x)=2xf(x)^2 $$
and hence $$\tag1f(x)\in\{0,1\}\quad\text{for }x\ne0. $$
The only continuous functions with this property are the constant functions
$$f(x)=0$$
and
$$f(x)=1. $$
Both are directly verified to indeed solve the original functional equation.



What if we drop continuity?

We still have $(1)$, but note that $f$ may jump discontinuously between $0$ and $1$.
Suppose $f(x_0)=0$ for at least one $x_0\ne 0$. Then with $y=x_0$, we get
$x_0f(x)=0 $
and hence 
$$ f(x)=0\quad\text{for all } x$$
as one solution (again).

So assume $f(x)=1$ for all $x\ne 0$. 
Then nothing can be said about $f(0)$, i.e., for any $c$,
$$f(x)=\begin{cases}c&amp;x=0\\1&amp;x\ne 0\end{cases} $$
is a solution, as one readily verifies.
Suppose $f(x) =0$ for all $x \neq 0.$ Since $f$ is continuous, so $f(0)=\lim_{x \to 0} f(x)=0$. Hence $f(x)=0$ for all $x \in \mathbb{R}.$

Similarly, the other case follows.