\begin {align*} y^{\prime } & =\sqrt {1-y^{2}}+x\\ y\left ( 0\right ) & =1 \end {align*}
\(f\left ( x,y\right ) =\sqrt {1-y^{2}}+x\) 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 does not apply.
In this case the ode has form \(y^{\prime }=f\left ( x,y\right ) \) and not \(y^{\prime }=f\left ( y\right ) \). So we can not just check if initial conditions satisfies the ode and use that as solution. If we did, we see that \(y\left ( x\right ) =1\) does satisfy the ode at \(x=0\) but this will be wrong solution. In this case we have to go ahead and solve the ode. In this case we will find that no general solution exists.