Find all functions $f : \mathbb{(0,\infty})\to\mathbb{R}$ such that $f(x+y) = xf(y)+ yf(x)$ 
Find all functions $f : \mathbb{(0,\infty)}\to\mathbb{R}$ such that $f(x+y) = xf(y)+ yf(x)$  if $f$ is continuous at $x=1$

This problem was looking quite easy at first but the domain of positive reals is posing me a problem. I couldn't plug in zero's for $x$ and $y$. I tried putting $x=y$ but the result $f(2x) = 2xf(x)$ couldn't be used as a recurrence relation $\infty$ times as that would yields $f(0)$ again. I've run out of ideas.
Please help. 
 A: If $f$ satisfies the given condition so does $cf$ for any constant $c$. Unless $f(1) = 0$  we might as well assume $f(1) = 1$. Some calculations:
$$f(2) = f(1 + 1) = f(1) + f(1) = 2$$
$$f(3) = f(1 + 2) = 2f(1) + f(2) = 4$$
$$f(4) = f(2 + 2) = 2f(2) + 2 f(2) = 8$$
$$f(4) = f(1 + 3) = 3f(1) + f(3) = 3 + 4 = 7$$
The last two equations are incompatible so we must have $f(1) = 0$.
Doing calculations like the above gives you $f(n) = 0$ and $f(2^{-n}) = 0$ for all natural numbers $n$. The additivity extends this to $f(r) = 0$ for all dyadic rationals $r$. This should give you that $f(x) = 0$ for all positive $x$ if you can exploit the continuity condition.
A: This is one possible continuation of your attempted solution.
By letting $\ y=x=z\ $ you found that $$ f(2z) = 2z f(z). \tag{1} $$
Let $\ y=2z,\ x=z\ $ to get $$ f(3z) = 2z f(z) + z f(2z). \tag{2} $$ 
Let $\ y=2z,\ x=2z\ $ to get $$ f(4z) = 4z f(2z). \tag{3} $$ 
Let $\ y=3z,\ x=z\ $ to get $$ f(4z) = 3z f(z) + z f(3z). \tag{4} $$ 
Now eliminate $\ f(2z),\ f(3z),\ f(4z)\ $ from the four linear equations to get
$$ (3-6z+2z^2)f(z) = 0. \tag{5} $$ There are two positive roots of the quadratic which implies that $\ f(z) = 0\ $ for any other values of $\ z.\ $ Use equation $(1)$ to prove $\ f(z) = 0\ $ for the positive roots also.
Notice that no continuity assumption or $\ f(0)\ $ was used to prove $\ f(z)=0\ \forall z.$
