Is it possible to solve this second order autonomous differential equation?

Multiply with $2x'$ and integrate $$ x'^2=C-\frac{2}{x}\implies x'=\pm\sqrt{C-\frac{2}{x}} $$ This now are two separable first order ODE.

Set $u=x'$ and re-insert into the original equation eliminating $x$, $\frac1x=\frac{C-u^2}2$ so that $$ u'=x''=\frac1{x^2}=\frac{(C-u^2)^2}4 $$ which should be solvable using separation and partial fraction decomposition to at least an implicit solution form.

$$x''(t)=\frac{1}{x^2(t)}$$ $$2x'x''=\frac{2x'}{x^2}$$ $$\left(\frac{dx}{dt}\right)^2=-\frac{2}{x}+c_1$$ With condition $-\frac{2}{x}+c_1\geq 0$ for all the following. $$\pm\sqrt{\frac{1}{-\frac{2}{x}+c_1}}dx=dt$$ $$\pm\int\sqrt{\frac{x}{x-2c_1}}dx=t+\text{constant}$$ I suppose that you can integrate for $t(x)$.

The inverse function $x(t)$ cannot be expressed with a finite number of elementary functions. For approximate solution use series or numerical calculus.

If some conditions where specified allowing to determine the constants of integration some simplification might occur.