How to solve the first-order ODE with initial condition?

Hint. The ODE can be rewritten as $$ \frac{dy}{dx}=\frac{y}{4\sqrt{xy}-x}, \tag{1} $$ which is homogeneous. The change of variable $y=xu$ transforms $(1)$ into a separable ODE: $$ x\frac{du}{dx}+u=\frac{xu}{4\sqrt{x^2u}-x}=\frac{u}{4\sqrt{u}-1}, \tag{2} $$ where I'm assuming $x>0$ because of the initial condition. Can you continue from here?

$$ ydx-\left( 4\sqrt{xy}-x\right) dy=0 $$ Multiplying the ode by integrating factor $\frac{1}{\sqrt{xy}}$ makes the ode exact \begin{align*} \frac{y}{\sqrt{xy}}dx-\left( 4-\frac{x}{\sqrt{xy}}\right) dy & =0\\ Mdx+Ndy & =0 \end{align*} Check: \begin{align*} \frac{\partial M}{\partial y} & =\frac{1}{2\sqrt{xy}}\\ \frac{\partial N}{\partial x} & =\frac{1}{2\sqrt{xy}} \end{align*} Let \begin{align} \frac{\partial\phi}{\partial x} & =M\tag{1}\\ \frac{\partial\phi}{\partial y} & =N\tag{2} \end{align} Integrating (1) gives \begin{equation} \phi=\int\frac{y}{\sqrt{xy}}dx=\frac{2xy}{\sqrt{xy}}+f\left( y\right) \tag{3} \end{equation} Taking derivative of the above and using (2) gives% \begin{align*} \frac{\partial\phi}{\partial y} & =N\\ \frac{x}{\sqrt{xy}}+f^{\prime}\left( y\right) & =-\left( 4-\frac{x} {\sqrt{xy}}\right) \\ f^{\prime}\left( y\right) & =-4\\ f & =-4y+c_{1} \end{align*} Plugging the above into (3) gives $$ \phi=\frac{2xy}{\sqrt{xy}}-4y+c_{1} $$ but $\phi$ is constant. Hence the above simplifies to (by combining constants) \begin{align*} \frac{2xy}{\sqrt{xy}}-4y & =c_{2}\\ 2\sqrt{xy}-4y & =c_{2}\\ \sqrt{xy}-2y & =c_{3} \end{align*} At $x=2,y=8$ \begin{align*} \sqrt{16}-2\left( 8\right) & =c_{3}\\ 4-16 & =c_{3}\\ c_{3} & =-12 \end{align*} Hence (4) becomes \begin{align*} \sqrt{xy}-2y & =-12\\ 2y-\sqrt{xy} & =12 \end{align*}

Hint

The equation writing $$y'=\frac{y}{4\sqrt{xy}-x}$$ let $y=x\,u^2$ and assuming $u\neq 0$, it becomes $$x\,u'+\frac {u (2u-1)}{4u-1 }=0$$ Switching variables $$\frac {x}{x'}+\frac {u (2u-1)}{4u-1 }=0 \implies \frac {x'}{x}=-\frac {4u-1 }{u (2u-1)}=-\frac{2}{2 u-1}-\frac{1}{u}$$ Integrating both sides is now simple. Inverse the solution, replace $u$ by its defintion and apply the condition.