\begin {align*} y^{\prime } & =\sqrt {1-y^{2}}\\ y\left ( 0\right ) & =1 \end {align*}
\(f\left ( x,y\right ) =\sqrt {1-y^{2}}\) is continuous in \(x\) everywhere. For \(y\) we want \(1-y^{2}\geq 0\) or \(y^{2}\leq 1\). The point \(y_{0}=1\) satisfies this. Now \(f_{y}=\frac {-2y}{2\sqrt {1-y^{2}}}\). We want \(1-y^{2}>1\) or \(y^{2}<1\). The point \(y_{0}\) does not satisfy this. Hence theory says nothing about uniqueness. Solution can be unique or not. When the ode has form \(y^{\prime }=f\left ( y\right ) \) we always check if IC satisfies the ode. In this case \(y\left ( x\right ) =1\) does satisfy the ode. So this means \(y\left ( x\right ) =1\) is solution. We do not need to solve by integration. But if we did, we will obtain the following
\begin {align*} \frac {dy}{\sqrt {1-y^{2}}} & =dx\\ \arcsin \left ( y\right ) & =x+c\\ y & =\sin \left ( x+c\right ) \end {align*}
At initial conditions the above gives \(1=\sin c\). Hence \(c=\frac {\pi }{2}\). Therefore solution is \(y=\sin \left ( x+\frac {\pi }{2}\right ) =\cos x\). So this is another solution that satisfies the ode. Solution is not unique.