I first note that a similar question was asked here: Calculating Radon Nikodym derivative, though the explicit steps used to calculate the derivative were not made clear.
Over measurable space $([0,1],\mathcal{B}_{[0,1]}), \mu:=\nu+\delta_{0}.$
$\forall A\in\mathcal{B}_{[0,1]}$ define $\nu(A):=|A|.$
Finally, $\delta_{0} $ is the dirac measure with criterion of measured set containing $0$ in order to evaluate to 1.
$\mu >>\nu$ is clear and therefore $d\nu/d\mu$ exists.
$d\nu/d\mu \overset{*}= \frac{\nu(dy)}{\mu(dy)}=\frac{|dy|}{|dy|+\delta_{0}(dy)} = 1/\frac{|dy|+\delta_{0}(dy)}{|dy|} =1/(1+\delta_{0}(dy)/|dy|)$. It is immediately apparent at this step, however, that this derivative will always be nonzero. Thus, I did something wrong in my calculation (and would like some help seeing where I went wrong). * is due to Notation when integrating with respect to a measure and is where I imagine I went wrong.
UPDATE:
It is clear that $\nu(A)=\int_{A}d\nu= \int_{A}\frac{d\nu}{d\mu}d\mu\Leftarrow \frac{d\nu}{d\mu}=\mathbf{1}_{\{X/0\}}$ to cancel out the dirac measure part in $\mu$ when it is not measure $0$, however, can the derivative be calculated from the differentials in some way or is it purely determined through the integral?
Your analysis was basically correct, and will be made rigorous below. At $$ \frac1{1+\frac{\delta_0(dy)}{|dy|}} $$ look what happens when $dy$ is around $0$. Then $|dy|$ is infinitely small,while $\delta_0(dy)=1$. Therefore, the above is $\frac1{1+\frac10}=\frac1{\infty}=0$, so the derivative is equal to $0$ at $y=0$, as expected. You were think $1/\text{stuff}$ could not be zero, but you forgot about stuff $=\infty$.
To get a differential characterization of $\frac{d\nu}{d\mu}$, it can shown that for $\mu$ a.e. $x$, you have $$ \frac{d\nu}{d\mu}(x)=\lim_{\epsilon\to 0}\frac{\nu((x-\epsilon,x+\epsilon)\cap [0,1])}{\mu((x-\epsilon,x+\epsilon)\cap [0,1])}. $$ This is a generalization of the Lebesgue differentiation theorem. I do not know what the standard reference for this is, but chapter 6.3 of these notes has a proof.
Now, when $x\neq 0$, then both the numerator and denominator are equal for all small enough $\epsilon$, so the limit is $1$. However, when $x=0$, the numerator will tend to zero while the denominator will tend to $1$. This shows that $$ \frac{d\nu}{d\mu}=1_{(0,1]}\qquad \mu \;a.e. $$