second derivative of a function equals a power of said function Is it possible to find an expression for $f(x)$ so that: 
$$f''(x) = (f(x))^{-2}$$
and 
$$f(0) = 1$$
I suspect that no such expression exists but i might be wrong.
Thanks in advance.
 A: Multiply by $f^\prime$ you get 
$$f^\prime f^{\prime\prime}=f^\prime f^{-2},$$
that is
$$\frac{d}{dx}\frac12(f^\prime)^2=\frac{d}{dx}(-f^{-1}),$$
Integrate both sides to get
$\frac12(f^\prime)^2=-f^{-1}+c.$ Hence $f'=\pm \sqrt{2c-2f^{-1}}$.
A: One can take the hard path and let 
$$f(x) = 1 + a_{1} x + a_{2} x^2 + a_{3} x^3 + a_{4} x^4 + \cdots,$$
since $f(0) = 1 = a_{0}$, such that 
\begin{align}
1 &= f^2 \, f'' = (1 + a_{1} x + a_{2} x^2 + a_{3} x^3 + a_{4} x^4 + \cdots)^2 \cdot (2 a_{2} + 6 a_{3} x + 12 a_{4} x^2 + \cdots) \\
&= (1 + 2 a_{1} x + (2a_{2} + a_{1}^2) x^2 + \cdots) \cdot (2 a_{2} + 6 a_{3} x + 12 a_{4} x^2 + \cdots) \\
&= 2 a_{2} + 2(2 a_{1} a_{2} + 3 a_{3} ) x + 2 (6 a_{4} + 6 a_{1} a_{2} + a_{1}^2 a_{2} + a_{2}^3) x^2 + \cdots.    
\end{align}
Equating the equations for the coefficients yields
\begin{align}
f(x) &= 1 + a_{1} x + \frac{x^2}{2!} - 2 a_{1} \, \frac{x^3}{3!} + \left( 6 a_{1}^2 - \frac{1}{2} \right) \, \frac{x^4}{4!} + a_{1} \, (13 - 24 a_{1}^2) \, \frac{x^5}{5!} \\
& \hspace{10mm} + ( 30 \, a_{1}^{2} \, (4 a_{1}^{2} - 5) + 1) \, \frac{x^6}{6!} + \cdots. 
\end{align}
Further terms may be obtained by considering more terms. Here $a_{1}$ is arbitrary unless another condition is applied.  It may be of interest that if $a_{1} = 0$, or $f'(0) = 0$, the the function $f(x)$ is an even function given by
$$f(x) = 1 + \frac{x^{2}}{2!} - \frac{1}{2} \, \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + \mathcal{O}\left(\frac{x^{8}}{8!}\right).$$
A: This is too long for a comment.
Using $$\dfrac{d^2x}{dy^2}=-\frac{\dfrac{d^2y}{dx^2}}{\left(\dfrac{dy}{dx}\right)^3}$$ the differential equation becomes $$\frac{x''}{(x')^3}+\frac{1}{y^2}=0\tag 1$$ and this one can be integrated leading to a very ugly result.
Reducing the order $(z=x')$, this leads to $$\frac{z'}{(z)^3}+\frac{1}{y^2}=0$$ leading to $$z=\pm\frac{\sqrt{y}}{\sqrt{2} \sqrt{c_1 y-1}}$$ Integrating once more $$x=\pm \frac{\sqrt{c_1} \sqrt{y} \sqrt{c_1 y-1}+\log \left(c_1 \sqrt{y}+\sqrt{c_1} \sqrt{c_1
   y-1}\right)}{\sqrt{2} c_1^{3/2}}+c_2$$
Try with Wolfram Alpha to see how frustrating is the result using $(1)$.
