A Calculus problem by Maximos Stratis

We divide both sides of the equation with $f^{3}(x)$ :
$\frac{f(x)f''(x)-2(f'(x))^{2}}{f^{3}(x)}=-e^{x}$
Now we multiply the numerator and the denominator of the first side with $f(x)$ :
$\frac{f^{2}(x)f''(x)-2f(x)(f'(x))^{2}}{f^{4}(x)}=-e^{x}$
The LHS is the derivative of $\frac{f'(x)}{f^{2}(x)}$ and the RHS is the derivative of $-e^{x}$ . Therefor:
$\frac{f'(x)}{f^{2}(x)}=-e^{x}+c_{1}$ , for some constant $c_{1}$
We substitute $x=0$ and we find that $c_{1}=0$ . So:
$\frac{f'(x)}{f^{2}(x)}=-e^{x}$
The LHS is the derivative of $-\frac{1}{f(x)}$ and the RHS is the derivative of $-e^{x}$ . Therefor:
$-\frac{1}{f(x)}=-e^{x}+c_{2}$ , for some constant $c_{2}$
We substitute $x=0$ and we find that $c_{2}=-1$ . So:
$\frac{1}{f(x)}=e^{x}+1\Rightarrow$
$f(x)=\frac{1}{e^{x}+1}$
and $f(ln3)=\frac{1}{e^{ln3}+1}=\frac{1}{4}=\boxed{0.25}$