How find all $f:\mathbb R\to\mathbb R$ such that $f\bigl(x\cdot f(y)\bigr)=y\cdot f(x)+kxy$ 
Let $k$ be a given real number. Find all the functions $f:\mathbb R\to\mathbb R$ such that
$$f\bigl(x\cdot f(y)\bigr)=y\cdot f(x)+kxy\,.$$

My try:
Let $x=y=0$, then
$$f(0)=0,$$
and $x=y=1$,
then
$$f\bigl(f(1)\bigr)=f(1)+k.$$
So let $f(1)=a$, then we have $f(a)=a+k$.
So I guess $f(x)=x+k$? Is it true? If this true, I can't prove it.
 A: Theorem. If $k<-\frac14$, no such function exists. If $k=-\frac14$, the only solution is $f(x)=\frac12x$. If $k> -\frac14$ and $k\ne 0$ there are exactly two solutions $f(x)=cx$ with $c=\frac{1\pm\sqrt{1+4k}}{2}$. If $k=0$ there are exactly two continuous solutions $f(x)=0$ and $f(x)=x$ and infinitely many non-continuous solutions (explicitly described in the proof below and most of them obtainable only using the Axiom of Choice).
Proof:
The function $f$ might be linear, $f(x)=cx$. In that case we obtain $c^2xy=cxy+kxy$ for all $x,y$, i.e. $c=\frac{1\pm\sqrt{1+4k}}{2}$ (only possible if $k\ge -\frac14$, of course). One checks that these are indeed solutions.
So from now on assume that $f$ is not linear.
As $f$ is not linear,
there exixts $x_0$ with $f(x_0)\ne -kx_0$.
Let $y_0=x_0f\left(\frac1{f(x_0)+kx_0}\right)$.
Then $$f(y_0) = f\Biggl(x_0f\left(\frac1{f(x_0)+kx_0}\right)\Biggr)=\frac1{f(x_0)+kx_0}\bigl(f(x_0)+kx_0\bigr)=1.$$
Then $$\tag1\label1 f(x)=y_0f(x)+ky_0 x\quad\text{for all }x.$$
If $y_0\ne1$, we can solve for $f(x)$ and obtain that $f(x)=\frac{ky_0}{1-y_0}\cdot x$ is linear, contrary to assumption.
Therefore $y_0=1$ and hence (with plugging in $x=1$ into \eqref{1}) we find $k=0$.
Starting all over again with $k=0$, the functional equation becomes $$\tag2\label2 f\bigl(xf(y)\bigr)=yf(x)\quad\text{for all }x,y$$ and has a more or less obvious solution
$$\tag3\label3 f(x)=\begin{cases}\tfrac1x&\text{if } x\ne 0\\0&\text{if }x=0.\end{cases}$$
More generally, let $\mathbb R=U\oplus V$ be any direct sum decomposition of $\mathbb R$ as $\mathbb Q$ vector space and let $g\colon \mathbb R\to\mathbb R$ be given by $g(u+v)=u-v$ for $u\in U, v\in V$.
Then
$$\tag4\label4 f(x)=\begin{cases}0&\text{if }x=0,\\
e^{g(\ln x)}&\text{if }x>0,
\\-f(-x)&\text{if }x<0\end{cases}$$
is a solution (check!) that is so to speak combined from $f(x)=x$ and $f(x)=\frac1x$.
Note that without invoking the Axiom of Choice we know only two such direct sum decompositions, namely those with one summand zero. These lead to the identity function ad \eqref{3}, respectively.
Can one show that all solutions for $k=0$ are of this form \eqref{4}? Not quite.
As $f$ is not linear, there exists $y$ with $f(y)\ne 0$. Then for all $t\in \mathbb R$ there exists $x$ with $t=xf(y)$. We find that $f\bigl(f(t)\bigr)=f\Bigl(f\bigl(xf(y)\bigr)\Bigr)=f\bigl(yf(x)\bigr)=xf(y)=t$, i.e. $f$ is an involution and especially $f$ is bijective. Then substituting $f(y)$ for $y$ in \eqref{2}, we obtain $$f(xy)=f(x)f(y)$$
and especially $f\restriction_{\mathbb R^\times}$ is a group automorphism.
We have $f(1)=1$ and $f(-1)=-1$ (because $f(-1)^2=f(1)$ and $f(-1)\ne f(1)$) and therefore $f(-x)=-f(x)$ for all $x$. Let
$$\hat f(x)=\operatorname{sgn}(x)|f(x)|.$$
Then one checks that $ \hat f(xy)=\hat f(x)\hat f(y)$ and $\hat f\left(\hat f(x)\right)=x$ and hence $\hat f$ is also a solution of the original functional equation, but with the additional property that $\hat f(x)>0$ iff  $x>0$.
Given $x>0$, we have $x\hat f(x)>0$ and can let $u=\sqrt{x\hat f(x)}$, $v=\frac xu$ so that $x=uv$. One checks that $\hat f\left(u^2\right)=u^2$ and hence $\hat f(u)=u$. Also, $\hat f(v)=\frac{\hat f(x)}{\hat f(u)}=\frac 1v$.
After taking logarithms, this corresponds exactly to the direct sum decomposition used above to construct \eqref{4}. We conclude that (for $k=0$) for any nonzero $f$ at least the sign-corrected function $\hat f$ is of the form \eqref{4}.
Remains to check how strange the signs of $f$ can be.
Since $f$ and $\hat f$ are both multiplicative, so is their quotient on $\mathbb R^\times$, which induces a group homomorphism from $\mathbb R^\times/\{\pm1\}\to\{\pm1\}$. And conversely, any $\hat f$ as given by \eqref{4} for a suitable direct sum decomposition of $\mathbb R$, together with a homomorphism $\mathbb R^\times/\{\pm1\}\to\{\pm1\}$ (essentially given by a subgroup of index $2$ in $\mathbb R_{>0}$ in the nontrivial case; again, without the Axiom of Choice we know only the trivial homomorphism) gives us a solution $f$ for the original functional equation. $_\square$
A: If $k=0,f=0$ is a solution.suppose there is such a function $f$ for $ k\neq 0$ and $1\in imf$(it's clear $f\neq 0$,we'll show that $1\in imf$),then there exist only one element such as $\alpha$ in $\mathbb{R}$ s,t $f(\alpha)=1$because in otherwise f isn't well-defined.now for arbitary $x\in\mathbb{R}$ we obtain $f(x)=\frac{kx\alpha}{1-\alpha}$.We must have $\alpha=\frac{-1\pm\sqrt{1+4k}}{2k}$(only one of these) because f satisfies required condition.it's clear that there exist a solution if $k\geq\frac{1}{4}$.
We shall show that $1\in imf$:$f(1.f(y))=f(f(y))=y.(f(1)+k)$,we consider $\alpha=f(\frac{1}{f(1)+k})$ and $f(\alpha)=1$.given what we have said so far,hence $f(x)=\frac{x}{\alpha}$($\alpha=\frac{-1+\sqrt{1+4k}}{2k}$or$\alpha=\frac{-1-\sqrt{1+4k}}{2k}$).
