A recent question asked for a solution to $f'(t)= k f^2(t)$. I could do this. In fact, it seems that a large class of similar differential equations $f'(t) = g(f(t))$ are not hard to solve.
This led me to consider $f''(t) = g(f(t))$.
I started with a specific example: $f''(t)= k f^2(t)$. I could solve this but I did so by just guessing that there would be a solution of the form $a x^b$. I found $f(x) = \frac{6}{k x^2}$ as a specific solution. Obviously, for this case and others of this class, translations of a solution as also solutions.
However, as $g$ becomes more complex, guesswork will probably fail me. Is there any good way to approach equations of this form?
Here is an attempt. Unfortunately, without knowing $g$ exactly, it is difficult to write a solution $f(t)$ to $$f''(t)=g\big(f(t)\big)$$ as a function of $t$ explicitly.
Write $x(t):=f(t)$ and $v(t):=f'(t)$. Then, you have $$v\,\frac{\text{d}v}{\text{d}x}=\frac{\text{d}^2x}{\text{d}t^2}=g(x)\,.$$ This implies $$\frac{1}{2}v^2=G(x)+c\,,$$ where $G$ is an antiderivative of $g$ and $c$ is some constant. That is, $$\frac{\text{d}x}{\text{d}t}=v=\pm\sqrt{2\,G(x)+2\,c}\,,$$ or $$\pm \sqrt{2}\,(t-\tau)=H_{c}(x)\,,$$ where $H_{c}$ is an antiderivative of $\dfrac{1}{\sqrt{G+c}}$ and $\tau$ is a constant. That is, by continuity, $$x=H_{c}^{-1}\big(+\sqrt{2}\,(t-\tau)\big)\text{ for }+\sqrt{2}\,(t-\tau)\in\text{Range}(H_c)\tag{1}$$ or $$x=H_{c}^{-1}\big(-\sqrt{2}\,(t-\tau)\big)\text{ for }-\sqrt{2}\,(t-\tau)\in\text{Range}(H_c)\,,\tag{2}$$ given that $H_c$ is injective.
An example with $g(u)=k\,u^2$ for some constant $k\neq 0$. Then, we can take $G(u)=\dfrac{k}{3}\,u^3$. That is, we can also take $$H_c(u)=-\int_{u}^\infty\,\frac{1}{\sqrt{\frac{k}{3}\,s^3+c}}\,\text{d}s\,.$$ I do not know the explicit form of $H_c$ for an arbitrary $c$, but $$H_0(u)=-\int_u^\infty\,\frac{1}{\sqrt{\frac{k}{3}\,s^3}}\,\text{d}s=-\frac{2\sqrt{3}}{\sqrt{k\,u}}\,.$$ Thus, the range of $H_0$ is $(-\infty,0)$. On $(-\infty,0)$, $$H_0^{-1}(z)=\frac{12}{k\,z^2}\text{ for }z<0\,.$$ Combining (1) and (2), a solution is given by $$x(t)=H_0^{-1}\big(-\sqrt{2}\,|t-\tau|\big)=\frac{6}{k\,(t-\tau)^2}$$ for all $t\in\mathbb{R}\setminus\{\tau\}$.
On the other hand, it is easier to solve for $f$ from $$f'(t)=g\big(f(t)\big)\,,$$ but you will still end up with some implicit relation. With the same notation as before, we have $$\frac{\text{d}x}{\text{d}t}=g(x)\,,$$ so $$\gamma(x)=t-\tau$$ for some constant $\tau$, and $\gamma$ is an antiderivative of $\dfrac{1}{g}$. Therefore, $$x=\gamma^{-1}(t-\tau)\text{ for }t-\tau\in\text{Range}(\gamma)\,,$$ provided that $\gamma$ is injective.
For $g(u)=k\,u^2$ with $k\neq 0$, we can take $$\gamma(u):=-\int_u^\infty\,\frac{1}{k\,s^2}\,\text{d}s=-\frac{1}{k\,u}\,.$$ That is, $$\gamma^{-1}(z)=-\frac{1}{k\,z}\text{ for }z\neq 0\,,$$ and we get a solution $$x(t)=-\frac{1}{k\,(t-\tau)}\text{ for all }t\in\mathbb{R}\setminus\{\tau\}\,.$$