how to calculate $d\Omega(f)$ here the question was to find $d \Omega(f)$ with : 
$$ \Omega : (E,[.]) \to (F,||.||) \\f \to -f'' +f^3$$
$
[f] = |f'(0)| + ||f''|| $
; $ ||f|| = Sup_{[0,1]}|f(x)| $
the answer is given to me like this : 
$ d\Omega(f)(h)= 3f^2h - h'' $ 
 , without any more details 
i want to know what is the role of h (i thought it's df at first) here , and how they calculated this answer 
note : 
it is true that $$lim_{[h]\to 0} {||\Omega(f+h) - \Omega(f) - d\Omega(f)(h)||\over [h]}=0$$ 
but i wanna know how they got that result and the role of (h) in the formula ? i struggled alot with this definition to understand it !
 A: I am assuming that $E=C^2([0,1])$ and $F=C([0,1])$.  Also, if $T$ is a linear transoformation, we gonna use the symbol $\langle T,x\rangle =T(x)$. Remember that the derivative of a function $F$ between two Banach spaces $X$ and $Y$  is a function $F':X\to L(X,Y)$ between $X$ and the set of all linear bounded transformations between $X$ and $Y$, $L(X,Y)$.
Write $T_1(f)=f^3$ and $\langle T_2, f\rangle =-f''$. As you can see, $T_2$ is a linear transformation, hence the derivative of $T_2$ is constant and equal to $T_2$ for all $f$, which implies that $$\langle T'_2(f),h\rangle=\langle T_2,h\rangle =-h''$$
Let's calculate $T'_1$. Fix some $h\in E$ and note that for $t$ real
\begin{eqnarray}\tag{1}
 \frac{T_1(f+th)(x)-T_1(f)(x)}{t} &=& \frac{(f(x)+th(x))^3-f(x)^3}{t}      \\
  \end{eqnarray}
For eaxh fixed $x\in [0,1]$, we can apply the mean value theorem to the function $y^3$ to conclude that $$\tag{2}\frac{(f(x)+th(x))^3-f(x)^3}{t}=\frac{3y(t,x)^2th(x)}{t}$$
where $y(t,x)\in [f(x),f(x)+th(x)]$ (the last interval is not oriented). Note that for each fixed $x$, if we let $t\to 0$, we conclude that $y(t,x)\to f(x)$, so, from $(1)$ and $(2)$ we have that $$\lim_{t\to 0}\frac{(f(x)+th(x))^3-f(x)^3}{t}=3f(x)^2h(x)$$
This is de Gateux derivative of $T_1$ with respect to the direction $h$. Take your conclusions from here.
