Does $f(x)$ exist so that $f(x)=xf'(x)$ except $f(x)=ax\,$? I want a function where its tangent line rotates around a fixed point and this is the only condition it must satisfy for that to happen.
 A: Here is how I would solve this, in order to handle all solutions at once.
Let $f$ be a differentiable solution.
Let $I_+ = (0,+\infty)$ and $I_- = (-\infty,0)$.
Let $f_{\pm}$ be the restriction of $f$ to $I_{\pm}$.
It is solution to the first order linear differential equation $y' = \frac{1}{x} y$ on the interval $I_{\pm}$, and by ordinary ODE theory, there exists a constant $c_{\pm}$ such that
$$
\forall x \in I_{\pm}, \quad f_{\pm}(x) = c_{\pm} e^{\int_{\pm 1}^x \frac{1}{t}dt} = c_{\pm} e^{\ln |x|} = c_{\pm}|x|.
$$
Note that on $I_{\pm}$, one has $|x| = \pm x$.
Let $\lambda_{\pm} = \pm c_{\pm}$, so that one can write
$$
\forall x \in I_{\pm}, \quad f_{\pm}(x) = \lambda_{\pm}x.
$$
By continuity, one has $f(x) = 0$.
Hence, there exist two constants $\lambda_+$ and $\lambda_-$ such that
$$
\forall x \in \Bbb R, \quad f(x) =
\begin{cases}
\lambda_+ x & \text{if } x \geqslant 0,\\
\lambda_- x & \text{if } x < 0.
\end{cases}
$$
Now, for $f$ to be differentiable, one needs to have $\lambda_+ = \lambda_-$, so that $f$ must be linear.
Conversely, any linear function is obviously a solution.
It follows that finally, the solutions are precisely the linear functions.
A: Surely $f(x)=0$ is a solution to the equation. When $f(x)$ is not identically zero, we have
$$
{f'\over f}(x)=\frac1x\Rightarrow\ln|f(x)|=\ln|x|+C\Rightarrow|f(x)|=e^C|x|
$$
Removing the absolute value, we see that $f(x)=Ax$ for some constant $A$ such that $|A|=e^C$.
