Solving $yy''=y^2y'+(y')^2$ with initial conditions 
Solve $\;\;yy''=y^2y'+(y')^2$ , $\quad\quad y(0)=\frac{-1}2,\quad y'(0)=1$

This equation has been asked several times on the site but my issue is finding the constants by knowing the initial conditions. Here is my approach,
Using the substitution $y'=z\quad , y''=\frac {dz}{dy}.z$
$$yzz'=y^2z+z^2\Rightarrow z=0\quad \text{or}\quad z'-\frac{z}y=y$$
From $z=0$ we have $y=c_1$ and since $y(0)=-\frac12$ then $y=-\frac12$ is a solution.
The solution to $ z'-\frac{z}y=y$ is $\quad y'=y(y+c)$. From here I want to use $y'(0)=1$ but there is no $x$ in the equation So I don't know how to find $c$. Continuation the solving,
$$\frac{dy}{y(y+c)}=dx\Rightarrow\quad x=\frac1c(\ln\left|\frac{y}{y+c}\right|)+d$$Where $c,d$ are constants. Writing the equation with respect to $y$,
$$e^{c(x-d)}=1-\frac c{y+c}\quad \Rightarrow \frac1{1-e^{c(x-d)}}=1+\frac{y}{c}\quad\Rightarrow y=c\left(\frac1{1-e^{c(x-d)}}-1\right)$$
Plugging in $x=0$ and $y=-\frac12$ will give an equation in two unknown $c,d$ so I can't find the values of the constants.
The book I'm reading from, only provide the final answer which is $y=\dfrac{3}{2(1-4e^{\frac32x})}$. But I don't know how to get it.

You have to insert all known quantities to determine $c$. Then you get
$$
1=-\frac12\left(-\frac12+c\right)\implies c=-\frac32.
$$
Then the constant in the second integration evaluates per
$$
0=-\frac23\ln\left|\frac14\right|+d.
$$
Also use
$$
\frac1{1-e^{-a}}-1=\frac{e^{-a}}{1-e^{-a}}=\frac1{e^a-1}
$$
