\begin {align*} y^{\prime } & =x\sqrt {y-3}\\ y\left ( 4\right ) & =3 \end {align*}
First we find the region where solution exists and is unique. Domain of \(f\left ( x,y\right ) =x\sqrt {y-3}\) is \(y-3\geq 0\) or \(y\geq 3\) since we do not want complex numbers and all \(x\) values. This shows solution exists. Domain of \(f_{y}=\frac {x}{2\sqrt {y-3}}\) is \(y>3\). Since point \(\left ( 4,3\right ) \) is not inside this domain (it can not be on the edge, it has to be fully inside), then theory do not apply. No guarantee that unique solution exist. Solving this gives\[ 2\sqrt {y-3}=\frac {1}{2}x^{2}+c \] At initial conditions\[ 0=8+c \] Hence \(c=-8\) and the solution becomes\begin {align*} 2\sqrt {y-3} & =\frac {1}{2}x^{2}-8\\ \sqrt {y-3} & =\frac {1}{4}x^{2}-4\\ y-3 & =\left ( \frac {1}{4}x^{2}-4\right ) ^{2}\\ y & =\left ( \frac {1}{4}x^{2}-4\right ) ^{2}+3 \end {align*}
Is this the only solution? Is this solution unique? No. By inspection we see that \(y=3\) is also a solution. Hence the solution exist but is not unique.