Taylor expansion $\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}$ I would like to use Taylor expansion around $y=0$ for that expression but i got stuck
$$\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}$$
Taylor expansion around $y=0$
$$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+o\left(y^3\right) $$
$$\displaystyle {1 \over y^2  + y^3}= {1 \over y^2 (1+y)} = {1 \over y^2}{1 \over 1+y} = {1 \over y^2}(1-y+\frac{y^2}{2} + o(y^2)).$$
$$\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}=\left(-\frac{y^2}{2}+\frac{y^3}{6}+o\left(y^3\right) \right) \cdot \left( {1 \over y^2}(1-y+\frac{y^2}{2} + o(y^2))\right)
= \left(-\frac{y^2}{2}+\frac{y^3}{6}+o\left(y^3\right) \right) \cdot \left( (\dfrac{1}{y^2}-\dfrac{1}{y}+\frac{1}{2} + o(1))\right)\\
= \left(-\frac{y^2}{2}+o\left(y^2\right) \right) \cdot \left( (\dfrac{1}{y^2}-\dfrac{1}{y}+\frac{1}{2} + o(1))\right)\\
=
$$
i'm stuck here 
 am i right 

No, your expansion is not right.
Although you wrote that $\frac{1}{y^2 + y^3} = y^{-2}(1 - y + y^2/2 + \cdots)$, you seem to have put all of that into the denominator. It shouldn't be there.
Instead, you should have
$$ (-y^2/2 + y^3/6 + \cdots ) y^{-2} (1 - y + y^2/2 + \cdots),$$
from which it is relatively easy to collect coefficients and describe the first coefficients of the Taylor expansion.
