Find all functions for which $\ x \cdot f(xy)+f(-y)=x \cdot f(x)$ Does anyone have idea for solution (for all non-zero numbers)?
$$\ x ≠ 0,y ≠ 0$$
$$\ f: R \setminus \{0\} → R$$
$$\ x \cdot f(xy)+f(-y)=x \cdot f(x)$$
Thanks!
 A: I can solve this completely given the additional assumption that $f$ is continuously differentiable on its domain. I suspect that it's possible to relax this to merely "continuous", but you'll probably need some sort of niceness assumption to get anywhere.
If we set $x=1$, the functional equation becomes
$$
f(y)+f(-y)=f(1) \, .
$$
So the equation is equivalent to
$$
x \, f(xy)-f(y)+f(1)=x \, f(x) \, ,
$$
which I will suggestively rewrite as
$$
x[f(xy)-f(x)]=f(y)-f(1) \, .
$$
Now, restrict to the case where $y$ is positive and set $y=1+\epsilon$. Since $f$ is differentiable, we can apply the mean value theorem to find $c_\epsilon \in [x, xy]$, $d_\epsilon \in [1, y]$ such that
$$
f(xy)-f(x)=x \epsilon \, f'(c_\epsilon)\\
f(y)-f(1)= \epsilon f'(d_\epsilon)
$$
(where we take the convention that interval endpoints are commutative: that is, if $a>b$, $[a,b]=[b,a]$).
So our equation becomes
$$
x^2\epsilon f'(c_\epsilon)=\epsilon f'(d_\epsilon) \, .
$$
But $f'$ is continuous by assumption, and we have $\lim_{\epsilon \to 0} c_\epsilon=x$, $\lim_{\epsilon \to 0} d_\epsilon=1$. So, passing to the limit,
$$
x^2 f'(x)=f'(1) \, ;
$$
that is, $f'(x)=\frac{C}{x^2}$ for all $x$ and some constant $C$, and so $f(x)=\frac{C}{x}+D$ for all $x$ and some constants $C,D$. 
Now, we can plug this $f$ back into the original functional equation to find which values of $C,D$ actually work. After a little bit of algebra, we see that the left-hand side of the equation reduces to $D+Dx$, while the right-hand side reduces to $C+Dx$. So we must have $C=D$.
That is, the continuously differentiable $f$ which satisfy the equation are precisely those of the form $f(x)=C(1+\frac{1}{x})$ for some constant $C$.
A: Partial Solution
$$ x \cdot f(xy)+f(-y)=x \cdot f(x)$$
Multiplying by $y$ we get
$$ xy \cdot f(xy)+yf(-y)=xy \cdot f(x) \,.$$
We define a new function $g(t):= tf(t) $.
Then, the functional equation becomes
$$g(xy)-g(-y)=yg(x) \,.$$
Thus
$$g(xy)=yg(x)+g(-y)$$
In particular, we get
$$\frac{g(xy)-g(-y)}{xy+y}=\frac{g(x)}{x+1} \,.$$
It is easy to see that $g(-1)=0$. Thus, with the extra assumption that $g(t)$ is differentiable at $-1$, by setting $x \to -1$, we get that $g$ is differentiable at $-y$ and $g'(-y)=g'(-1)$. Let $g'(-1)=C$, then $g(t)= Ct+D$ for some constants $C,D$, and the functional equation Yields $c=d$.
A: I guess there doesn't lies any solution to this one .
Replacing all y's by x. We get x∗f(x^2)+f(−x)=x∗f(x)....(1)in this equation if we put x=1,we get 
f(-1)=0 .So let us consider there lies a function Q(x) such that , f(x)=(x+1)Q(x). Replacing the f(x) in equation (1) we get one more equation, if we consider the lading coefficient of Q(x) be a1, and it be of degree n.....we see that n=0.
so there doesn't lie a function as Q(x). we get f(x)=(x+1);which actually does't satisfy equation 1.
so we conclude that f(x) is a constant function and is identically equal to zero.
ALTERNATIVELY
considering the given functional equation if we  differentiate both sides with respect to x  we get  
