Cauchy problem \begin{cases} \frac{1}{y'}-x=y x^2, \ \ \ x>0, y>-\frac{1}{x} \\ y(1)=0 \\ \end{cases} Discuss the following Cauchy problem.
\begin{cases}
\frac{1}{y'}-x=y x^2, \ \ \ x>0, y>-\frac{1}{x} \\
y(1)=0 \\
\end{cases}
My approach:
$$y' (x+y x^2)=1\Rightarrow \ y' (1+y x)=\frac{1}{x}$$
By integration:
$$y+\frac{1}{2}\int [y^2]' x dx=\ln x + C$$
Integrating by part
$$y+\frac{1}{2}\left(y^2x-\int y^2 dx\right)=\ln x + C$$
And here I stop! Any suggestions please?

$$\frac{dx}{dy}-x=yx^2$$
using, $x=\frac{1}{t}$ , so that $\frac{dx}{dy}=-\frac{1}{t^2}\frac{dt}{dy}$
Using these two we get, 
$$-\frac{1}{t^2}\frac{dt}{dy}-\frac{1}{t}=\frac{y}{t^2}$$
Which gives, 
$$\frac{dt}{dy}+t=-y$$
Now there are two possible ways to solve this one with $-(t+y)=z$ substitution and the other being realizing the fact it's a first order differential equation.
Though , I am going for the former. You can try the latter.
$$\frac{dt}{dy}=-(t+y)$$
put $t+y=-z$ , so that $\frac{dt}{dy}+1=-\frac{dz}{dy}$
Which gives, 
$$-1-\frac{dz}{dy}=-z$$
$$\frac{dz}{dy}=z-1$$
Now it's a really simple separable  differential equation yielding,
$$\ln|z-1|=y+c$$
Now it's just a matter of reverse substitution but be careful with the signs, domains and ranges. :)
Most seemingly , $\frac{1}{x}+y+1=ce^y$
