Question on nonlinear IVP 
$$y' (x) = (1 - 2x) y^2, \qquad y(0) = y_0$$
(A) Show that this initial value problem has a unique solution for any choice of $y_0$.
(B) How does the interval of definition of the solution depend on $y_0$?


Could you explain the procedure to answer those questions? Please.
I am knew to this site. Hopefully It will end my search.
 A: You can solve this by separation of variables, provided $y$ is not $0$, since you have to divide by $y$ to separate variables.  If I'm not mistaken, you should get
$$
y = \frac{1}{x^2-x+C}.
$$
The function $y=$ for all $x$, is also a solution, as you can check by substitution.
As for "intervals of definition", notice that the quadratic equation $x^2-x+C=0$ has no solutions if $C<-1/4$.  If $C>-1/4$, it has two solutions, and so you get vertical asymptotes in the solution.  So you might say the interval on the $x$-axis between $0$ and the nearest vertical asymptote to $0$ is the "interval of definition".  But really, the two vertical asymptotes break the line into three intervals.  You get solutions on each interval.  But only one of the intervals contains $0$.  So on that interval, the initial condition specifying $y(0)$ determines the solution; on the other two intervals it doesn't.
Since $y(0)=1/C$, we get $C= 1/y(0)$.
But which of the three intervals contains $0$?  That depends on $C$ and hence on $y(0)$.  The solution to the quadratic equation is
$$
x=\frac 1 2 \pm \sqrt{\frac 1 4 + C\ {}}.
$$
If $C$ is just a little bit bigger than $-1/4$, then both of these numbers are positive, so $0$ is to the left of both of them.  If $C>0$ then $0$ is between the points where the vertical asymptotes are.  If $C=0$ exactly, then one of the vertical asymptotes is at $0$ and there will be no solution that satisfies an initial condition saying $y(0)={}$ some finite number.
What about uniqueness?  For that, I might start by looking at the Picard–Lindelöf theorem.
