How can we find whether this problem has unique, none, infinitely many solutions? Everytime I encounter this kind of question, I got stuck. Please help me if anyone knows how to solve and teach me how to solve.

This is what I did basically,
$$f(x,y)=(1-y^2)^{1/2}$$
$$f(y)=-y/(1-y^2)^{-1/2}$$  where $y \ne {-1,1}$
Then I got stuck. 
 A: $$
\int \frac{1}{\sqrt{1-y^2}}dy = \int dx \Rightarrow \arcsin (y) = x + C
$$
$$
y = \sin(x+C) \Rightarrow \begin{cases}|b|> 1 \rightarrow \mbox{No Solution} \\ |b| \leq 1 \rightarrow \mbox{Unique Solution}\end{cases} 
$$
A: The given ODE is $x$-free. This implies that the set of solution curves is invariant under horizontal translation, hence the number of solutions of an IVP stipulating $y(a)=b$ does not depend on $a$.
It is obvious that there are no solutions if $|b|>1$. Therefore it remains to look at the interval $-1\leq b\leq 1$. 
Consider the right side
$f(x,y):=\sqrt{1-y^2}$ of the given ODE. When $|b|<1$ then
$${\partial f\over\partial y}={-y\over\sqrt{1-y^2}}$$
is bounded in a neighborhood of $(a,b)$. In this case the general existence and uniqueness (GEU)  theorem says that there is an interval $\ ]a-h, a+h[\ $ such that there is exactly one solution
$$\phi:\quad]a-h,a+h[\ \to{\mathbb R}$$
satisfying $\phi(a)=b$.
The case $b=1$ is special ($b=-1$ is exactly analogue). Here the basic "technical assumption" of the GEU theorem is violated, and trouble is in the offing. Note that the "ordinary solutions" of the given ODE are sine arcs of the form
$$y=\sin(x-c)\qquad\left(c-{\pi\over2}<x<c+{\pi\over2}\right)\ .$$
Here the solution $y=\sin\bigl(x+{\pi\over2}\bigr)$ living on the interval $\ ]-\pi,0[\ $ can be smoothly extended to the point $(0,1)$. Furthermore there is the "special solution" $y_1(x):\equiv1$ going through this point. Out of this material we can fabricate essentially two different solutions through the initial point $(0,1)$, namely $x\mapsto y_1(x)$ and 
$$x\mapsto y_2(x)=\left\{\eqalign{&-1\qquad\qquad\qquad(x<-\pi)\cr 
&\sin\bigl(x+{\pi\over2}\bigr)\qquad(-\pi\leq x\leq 0) \cr &1\qquad \qquad\qquad(x>0)\ .\cr}\right.$$
Any solution through $(0,1)$ coincides in a neighborhood of $x=0$ with one of $y_1$ or $y_2$.
