Can anyone solve this hard differential equation involving a derivative squared?

First, write $h(t)=H_0+f(t)$ to get a slightly simpler equation $$ \frac{2gA_O^2}{A_O^2 - A_ G^2}f(t)+\frac{A_O^2}{A_G^2}\bigg(\frac{df(t)}{dt}\bigg)^2=0. $$ Writing $r=A_O/A_G$, we can simplify the constants and find $$ \frac{2g}{r^2 - 1}f(t)+\bigg(\frac{df(t)}{dt}\bigg)^2=0. $$ If $f'(t)=0$, the equation tells you $f(t)=0$.

I'll assume $r>1$ so that the equation forces $f\leq0$. This holds as $A_0>A_G$. The other case will be similar but I will stick to this choice as in the updated question.

If $f'(t)\neq0$, you get $$ f'(t) = \pm\sqrt{-\frac{2g}{r^2 - 1}f(t)} $$ or $$ \frac{df}{-\sqrt{\frac{2g}{r^2 - 1}f}} = \pm dt. $$ Integrating gives $$ \sqrt{\frac{r^2 - 1}{8g}} \sqrt{-f} = t_0\pm t $$ for some $t_0$, so $$ f(t) = -\frac{8g}{r^2 - 1} (t_0\pm t)^2. $$ As the time difference is squared anyway, you can rewrite this general solution as $$ f(t) = -\frac{8g}{r^2 - 1} (t-t_0)^2. $$ This gives you two kinds of solutions, and for $r<1$ you will get something similar.

Be careful with solutions when $f=0$ (which is at $t=t_0$). There the solution fails to be unique because the function can stop at the zero level. For example, $$ f(t) = \begin{cases} -\frac{8g}{r^2 - 1} t^2,&t<0\\ 0,&0\leq t\leq 1\\ -\frac{8g}{r^2 - 1} (t-1)^2,&t>1 \end{cases} $$ is a solution.

Your definition of $H_0$ requires that $f(0)=0$, as $f=0\iff h=H_0$. With your update I think you mean $h(0)=0$ instead of $h(0)=H_0$.

Let us then see what happens with your added assumption that $h(t)\leq H_0$, $h$ is strictly increasing and $\lim_{t\to\infty}h(t)=H_0$. With my notation this gives $f(t)\leq0$, $f'(t)\geq0$, and $\lim_{t\to\infty}f(t)=0$. The updated initial condition seems to be $f(0)=-H_0$. Let's start with $$ f(t) = -\frac{8g}{r^2 - 1} (t-t_0)^2. $$ Putting in the initial condition gives $$ -H_0 = f(0) = -\frac{8g}{r^2 - 1} t_0^2, $$ so $$ f(t) = -a (t-\sqrt{H_0/a})^2, $$ where $a=\frac{8g}{r^2 - 1}$. As $f$ should be increasing, we have $$ f(t) = \begin{cases} -a (t-\sqrt{H_0/a})^2 ,& 0\leq t\leq \sqrt{H_0/a}\\ 0 ,& t>\sqrt{H_0/a}. \end{cases} $$ The function is increasing but not strictly increasing; it reaches it's final value in finite time. Without the increasing assumption the solution can decide to "bounce back down" starting at any time after $t_0=\sqrt{H_0/a}$.

You could put it in the form $$\left(y’\right)^2 = ky $$ where $y =h-H_0$ and $k=-\lambda/\mu$.

With assumptions about various signs, you might even go for

$$y’=cy^{1/2}$$