Solution to differential equation $f^2(x) f''(x) = -x$ on [0,1] I'd like to solve a differential equation $$ f^2(x) f''(x)=-x $$ where $f(x)$ is defined on $[0,1]$ and has a boundary condition $f(0)=f(1)=0$.
I somehow found out that the solution is fairly close to $f(x) = x^{1/3} \phi^{2/3}(\Phi^{-1}(1-x))$ where $\phi$ and $\Phi$ are pdf and cdf of a standard normal distribution, but it fails to solve the differential equation exactly.

Thank for all comments!
Based on the solution structure of Emden–Fowler Equation, I was able to identify the values of constants that satisfy the boundary conditions. The followings are the details: 
Define
\begin{equation}
 Z_R(\tau) \triangleq \sqrt{3} J_{1/3}(\tau) - Y_{1/3}(\tau)
 , \quad
 Z_L(\tau) \triangleq - \frac{2}{\pi} K_{1/3}(\tau)
\end{equation}
where $J, Y, K$ are Bessel functions.
Further define
\begin{equation}
 \bar{\tau} \triangleq \inf\{ \tau > 0; Z_R(\tau) = 0 \} \approx 2.3834
 , \quad
 a \triangleq \frac{1}{ \bar{\tau}^{4/3} Z_R'(\bar{\tau})^2 } \approx 0.2910
 , \quad
 b \triangleq a \left( \frac{9}{2} \right)^{1/3} \approx 0.1763.
\end{equation}
Then, the solution curve $\{ (x, f(x)) \}_{x \in [0,1]}$ is characterized by
\begin{equation}
 \left\{ \left( x_R(\tau), y_R(\tau) \right) \right\}_{\tau \in [0, \bar{\tau}]} \bigcup \left\{  \left( x_L(\tau), y_L(\tau) \right) \right\}_{\tau \in [0, \infty]} 
\end{equation}
where
\begin{equation}
 x_R(\tau) \triangleq a \tau^{-2/3}\left[ \left( \tau Z_R'(\tau) + \frac{1}{3} Z_R(\tau) \right)^2 + \tau^2 Z_R(\tau)^2 \right]
 , \quad
 y_R(\tau) \triangleq b \tau^{2/3} Z_R(\tau)^2.
\end{equation}
\begin{equation}
 x_L(\tau) \triangleq a \tau^{-2/3}\left[ \left( \tau Z_L'(\tau) + \frac{1}{3} Z_L(\tau) \right)^2 - \tau^2 Z_L(\tau)^2 \right]
 , \quad
 y_L(\tau) \triangleq b \tau^{2/3} Z_L(\tau)^2.
\end{equation}

In addition to this analytic solution, I also obtained a numerical solution by repeatedly computing
$$ f_{k+1}(x) \gets \left[ \left( f_k(x-2h) + f_k(x+2h) \right) + 4 \left(f_k(x-h)+f_k(x+h)\right) + \frac{8 x h^2}{f_k^2(x)} \right] \big/ 10 $$
on the grid $x \in \{2h,3h,\ldots,1-3h,1-2h\}$ for small $h$ with an initialization $f_0(x) \triangleq 0.5(1-(1-2x)^2)$.
The following figure shows these solutions:

 A: I tried the following approach.  Put $y=f(x)$ and $t=2x-1$ so the differential equation becomes $y^2\ddot{y}=-(t+1)/8$ with boundary conditions $y=0$ at $t=\pm 1$.  We can then write $y=\sum_ia_it^i$.  The differential equation gives a recurrence relation expressing all the coefficients $a_i$ in terms of $a_0$ and $a_1$.  We can then truncate the power series to a given order $d$ and solve numerically for the boundary conditions.  This seems to work in a well-behaved way, with good convergence at the endpoints and a result that is stable when we increase $d$.  It looks like $a_0=0.450$ and $a_1=0.120$ to $3$ decimal places.  Maple code is as follows:
with(plots):
Digits := 50:
d := 50:
y := add(a[i] * t^i,i=0..d):
sol0 := solve([coeffs(rem(expand(y^2 * diff(y,t,t) + (t+1)/8),t^(d-1),t),t)] 
              {seq(a[i],i=2..d)}):
y0 := expand(subs(sol0,y)):
sol1 := fsolve({subs(t= 1,y0),subs(t=-1,y0)},{a[0]=0.45,a[1]=0.1}):
aa[d] := subs(sol1,[a[0],a[1]]);
y1 := subs(sol1,y0);
y1x := subs(t = 2*x-1,y1):
Phi := unapply((1 + erf(x))/2,x):
phi := unapply(diff(Phi(x),x),x):
display(
 plot(y1x,x=0..1,colour=red),
 plot(x^(1/3) * phi(RootOf(1-x-Phi(_Z)))^(2/3),x=0..1,colour=blue)
);

This generates the following picture.  The power series solution is in red and the function $x^{1/3}\phi(\Phi^{-1}(1-x))^{2/3}$ is in blue.

The coefficients $a_{2i}$ lie on a nice smooth curve, and the coefficients $a_{2i+1}$ lie on a similar curve shifted down slightly.  Logs of the absolute values can be displayed as follows:
display(
  listplot([seq(log(-coeff(y1,t,2*i)),i=3..(d-1)/2)],style=point,colour=red),
  listplot([seq(log(-coeff(y1,t,2*i+1)),i=3..(d-1)/2)],style=point,colour=blue)
);

One could probably get further by finding an exact or approximate formula for these curves.

A: Surprisingly, this case of the Emden-Fowler equation is explicitly solvable:
see formula (2.3.27) in A. Polyanin and V. Zaitsev, Handbook of exact solutions
of ordinary differential equations, Chapman & Hill, 2003.
I copy the formula, without verifying it. Let
$$Z=C_1J_{1/3}(\tau)+C_2Y_{1/3}(\tau),$$
or
$$Z=C_1I_{1/3}(\tau)+C_2K_{1/3}(\tau),$$
where $J,Y$ are Bessel and $I$, $K$ are modified Bessel functions.
Then
$$x=a\tau^{-2/3}[(\tau Z^\prime+(1/3)Z)^2\pm\tau^2Z^2],\quad y=b\tau^{2/3}Z^2$$
satisfy $d^2y/dx^2=Axy^{-2}$ with $A=-(9/2)(b/a)^3.$
For the $+$ sign in $\pm$ take the first formula for $Z$, and for the $-$ the second one.
Remark. Emden-Fowler equation appears for the first time in the famous book by R. Emden, Gaskugeln (1907) and since then frequently arises in the study of stars and black holes.
