= Solution
For real solutions with $x>0$, the square root requires $y\ge0$. First retain the solution $y\equiv0$, which would be lost by dividing by a square root of $y$.
For $y>0$, use the <homogeneous first-order differential equation> substitution $v=\sqrt{y/x}>0$, so $y=xv^2$, $\sqrt{xy}=xv$ and $y'=v^2+2xvv'$. The equation becomes
$$
(1+v)(v^2+2xvv')=v^2,
\qquad\frac{2(1+v)}{v^2}\,dv=-\frac{dx}{x}.
$$
This is now a <separable differential equation>. Integration gives $-2/v+2\ln v=-\ln x+C$. Since $xv^2=y$, the positive solutions satisfy
$$
\boxed{\ln y-2\sqrt{\frac xy}=C\quad(y>0),\qquad\text{together with }y\equiv0.}
$$
The implicit family is sufficient here because only parts (a) and (b) request an explicit expression. It also shows why the nonnegative choice of $v$ matters: $\sqrt{xy}$ denotes the positive square root.
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