I've found this derivative on a textbook
$\dfrac{d(c_{t+1}/c_t)}{d(\dfrac{\gamma}{c_t}/\dfrac{1-\gamma}{c_{t+1}})}=\dfrac{1-\gamma}{\gamma} \dfrac{d(c_{t+1}/c_t)}{d(c_{t+1}/c_t)}=\dfrac{1-\gamma}{\gamma}$
I would like to understand the first passage. Was $\dfrac{1-\gamma}{\gamma}$ just brought out of the $d()$ at the denominator?
A more rigorous way to do this is as follows. Define a new variable: $$S\equiv\frac{\dfrac{\gamma}{c_t}}{\dfrac{1-\gamma}{c_{t+1}}}.$$ By simple rearrangement, you can compute that $$\frac{c_{t+1}}{c_t}=\frac{1-\gamma}{\gamma}\times S.$$ Hence, if you imagine $c_{t+1}/c_t$ as a function of $S$, you have that $$\frac{\mathrm{d}\left(\dfrac{c_{t+1}}{c_t}\right)}{\mathrm d S}=\frac{1-\gamma}{\gamma},$$ since $c_{t+1}/c_t$ is just a linear function of $S$ with coefficient $(1-\gamma)/\gamma$.
In your original question, “bringing out $(1-\gamma)/\gamma$ out of the denominator” is an informal operation, since what appears there is, in fact, a differential operator and, rigorously speaking, there is no “denominator;” it's just a conventional notation. However, the informal trick of “bringing out $(1-\gamma)/\gamma$ out of the denominator” does work in most cases, which serves as a justification for denoting the differential operation as though it were a fraction.
In general, you may want to use the implicit function theorem to take the derivative of one quantity with respect to another if the functional relationship between these two quantities is complicated.
You must refer to the definition of derivative (coefficient brings out from derivation) and notice that in the fraction you mentioned above (in fact a notation just!) the denominator is an independent object and nominator -usually- is a function of it. finally Do derivative to obtain final result!
Thank you to the answerer. I've been thinking on it, I could have used also the inverse function theorem.
$\dfrac{d(c_{t+1}/c_t)}{d(\dfrac{\gamma}{c_t}/\dfrac{1-\gamma}{c_{t+1}})}\quad (1)$
is the inverse of
$\dfrac{d(\dfrac{\gamma}{c_t}/\dfrac{1-\gamma}{c_{t+1}})}{d(c_{t+1}/c_t)}= \dfrac{d (\dfrac{c_{t+1}}{c_t} \cdot \dfrac{\gamma}{1-\gamma})}{d(c_{t+1}/c_t)}=\dfrac{\gamma}{1-\gamma} \quad (2) $
So since the result of (1) is equal to the reciprocal of the result of (2), the derivative will be
$\dfrac{1-\gamma}{\gamma}$