Does a "Newton differential equation" always have a solution? One reason why the iteration
$$x_{n+1}=x_n-\tan\;x_n$$
converges quickly for appropriate starting values is that this is nothing more than the Newton-Raphson iteration for $\sin\;x$.
This got me thinking: given some arbitrary function $g(x)$, is there always a function $f(x)$ such that
$$\frac{f(x)}{f^\prime(x)}=g(x)$$
or are there restrictions on the nature of $g(x)$ so that the differential equation has a solution?
 A: The differential equation you have is an example of a first-order separable differential equation.
If you write $y = f(x)$, then you have 
$$
\frac{y}{\frac{dy}{dx}} = g(x)
$$
Writing this in terms of differentials (treat $\frac{dy}{dx}$ as a fraction) so that all the $y$'s are on one side and all the $x$'s are on the other gives
$$
\frac{dx}{g(x)} = \frac{dy}{y}
$$
Integrating both sides yields
$$
\int \frac{dx}{g(x)} + C = \ln |y|
$$
where $C$ is an arbitrary constant.  Then
$$
\exp(\int \frac{dx}{g(x)} + C) = |y|
$$
and so
$$
y = \pm \exp(\int \frac{dx}{g(x)} + C)
$$
This works as long as $\frac{1}{g(x)}$ has an antiderivative.  For this purpose, it is sufficient to have $g$ continuous and not everywhere zero.
A: Certainly not for "arbitrary" functions, as functions of the form $\frac{f}{f'}$ satisfy properties that not all functions satisfy.  In fact, in any open set where $g$ is nonzero, $\frac{1}{g}=\log(|f|)'$, so $g$ must be the reciprocal of a derivative of a function.  This also leads to a way to find such $f$ in many cases, namely by integrating the last equation to get $\log(|f(x)|)=\int_a^x\frac{dt}{g(t)}+C$ provided the integral exists, in which case $f(x)=c\cdot\exp\left(\int_a^x\frac{dt}{g(t)}\right)$.
For example, attempting this with $g(x)=\tan(x)$ leads to integrating $\cot(x)$ and ending up with $f(x)=c\cdot\sin(x)$.
