Examine whether the initial value problem $\frac{dy}{dx}$ = $x\sqrt{y-3}$ , $y(4)=3$ has unique solution or more than one solution or no solution.

Question

My attempt

Let $f(x, y)=x \sqrt{y-3}$ then $f_y (x, y)= \frac{x}{2\sqrt{y-3}}$

We see that $f_y (x, y)$ does not exist at (4,3). We can't conclude whether the solution(s) exists or not.

Solving we get $y=3+(\frac{x²}{4}+c)²$

Using initial condition we get, $y=3+(\frac{x²}{4}-4)²$

But in my book it is given that $y =$ 3 is also a solution (but how to obtain this solution?)

I can't understand what is this? And what is missing solution? Please give me some hints. Thank you