Has this second-order nonlinear ODE been studied before or been given a known name that I can look up for further study or investigation?
$$ y’’ \sqrt{y}+f(x)=0, $$
in single real variable $x$, with $y=y(x)$, an arbitrary function $f(x)$, and with the double primes indicating second derivative with respect to $x$?
On
For the case where $f(x)$ is a constant $a$, first switch variable and write $$-\frac {x''}{[x']^3 }\sqrt y=a$$ Now, reduction of order $p=x'$ $$\frac {p'}{p^3}=-\frac a{\sqrt y}\implies p=\pm \frac{1}{\sqrt{4 a \sqrt{y}+ c_1}}$$ $$x'=\pm \frac{1}{\sqrt{4 a \sqrt{y}+ c_1}}\implies x+c_2=\pm\frac{\left(2 a \sqrt{y}-c_1\right) \sqrt{4 a \sqrt{y}+c_1}}{6 a^2}$$ Squaring and using $z=\sqrt{y}$ $$16 a^3 z^3-12 a^2 c_1 z^2+c_1^3=36a^4(x+c_2)^2 \tag 1$$
Assuming $c_2=0$ and $z(0)=0$ imply $c_1=0$ which gives in the real domain $$z=\sqrt[3]{\frac{9 a }{4}x^2} \implies y=\Bigg[\frac{9 a }{4}\Bigg]^{\frac 23} \, x^{\frac 43}$$
Otherwise, we just need to solve the cubic equation $(1)$ in $z$ .
Let us write the equation in the form \begin{align} y' ={}& v\\ v' ={}& \frac{f}{\sqrt{y}} \end{align} and as mentioned in the comments we'll consider the case $f\equiv const\ne 0$. We write $$ vv' = \frac{y'}{\sqrt{y}}f $$ which, due to a constant $f$, has the first integral $$ \frac{1}{2}v^2 = 2y^\frac12 f + c $$ where $c$ is a constant of integration. Let us define $$2y^\frac12 f + c = \frac12 u^2$$ from which we infer $$y=\frac{1}{4f^2}(\frac12 u^2 - c)^2\implies y'=\frac{1}{2f^2}(\frac12 u^2 - c)uu' $$ and so $$(y')^2=v^2=u^2$$ naturally leads to the two branches $$(\frac12 u^2-c)u'=\pm 2f^2.$$ Integrating yields $$\frac16 u^3 - cu = b\pm2f^2x$$ where $b$ is another constant of integration. Taking $c=0$ there is an obvious solution (well... two obvious solutions as it has two branches... or even six if we consider the complex branches of $\cdot^\frac13$) $$u=(6(b\pm2f^2x))^\frac13\implies y=\frac{1}{16f^2}(6(b\pm2f^2x))^\frac43.$$ Setting $b=0$ yields $$y=\big(\frac{9}{4}f\big)^\frac{2}{3} x^\frac{4}{3} .$$ However let's go for something a bit harder with $c\ne 0$ (in particular with $c>0$). We'll tidy things up a bit and write $p=6c$ and $q\equiv q(x)=-6(b\pm2f^2x)$. Thus we have that $$u^3-pu+q=0.$$ Okay, now cubics are gross but lets plow ahead anyway. I will assume from now on that $\big(\frac{q}{2}\big)^2\le\big(\frac{p}{3}\big)^3$. Note this implicitly assumes $p>0$ (which in turn implies $c>0$). I'll leave you to consider the other cases. There is a cubic formula but I don't like it and I prefer to write solutions in terms of trig or hyperbolic trig functions. Let us write $u=r\cos(\frac{\theta}{3})$, $r>0$. Applying the triple angle formula, an obvious choice is for $r$ to satisfy $\frac34 r^3 = pr$. Or equivalently, $$r = 2\big(\frac{p}{3}\big)^\frac12.$$ This leaves us with $$\frac{1}{4}r^3 \cos(\theta)+q=0\implies \cos(\theta) = -\frac{\frac{q}{2}}{\big(\frac{p}{3}\big)^\frac32}.$$ Note that my earlier assumption ensures that $|\cos (\theta)|\le 1$. Therefore we can write down all three solutions to the cubic via $$u=2\big(\frac{p}{3}\big)^\frac12\cos\big(\frac{\theta + 2n\pi}{3}\big)$$ where $n\in\mathbb{Z}$ and $$\theta = \arccos\Big(-\frac{\frac{q}{2}}{\big(\frac{p}{3}\big)^\frac32}\Big) = \arccos\Big(\frac{3(b\pm2f^2x)}{(2c)^\frac32}\Big).$$ Recalling our formulae for $y$ and $p$, $$y = \frac{c^2}{4f^2}\Big(4\cos^2\big(\frac{\theta + 2n\pi}{3}\big) - 1\Big)^2.$$ Hopefully I haven't made any mistakes. Note that there are implicitly 6 different branches contained in this solution.