Proving $x\equiv 2^{11} \mod{97}$ In $\mathbb{N}$ we suppose the equation $(F): x^{35} \equiv 2 \mod{97}$, and $(97,x)=1$.
How could I prove that $x\equiv 2^{11} \mod{97}$ ?

Here we look for relations ('tricks') that will solve the problem without using Fermat's Little Theorem.
We interpret the OP's question as a 'how to calculate' ${2^{11}}^{35} \pmod{97}$.
Well $2^{11} \equiv 2^8 \times 2^3 \equiv 62 \times 8 \equiv 11 \pmod{97}$ so we've found our first relation,
$\tag 1 {2}^{11} {11}^{-1} \equiv 1  \pmod{97}$
We suspect that progress can be made by calculating ${11}^{-1} \pmod{97}$, i.e. solving the equation 
$\quad 11x \equiv 1 \pmod{97}$ 
This can be done by hand using an algorithm discussed here. Proceeding,
Multiply by $ 9 $:
$\quad 99 x \equiv 9 \pmod{97}$
$\quad 2 x \equiv 9 \pmod{97}$
Multiply by $ 48 $:
$\quad 96 x \equiv 432 \pmod{97}$
$\quad -1 x \equiv 44 \pmod{97}$
$\quad 1 x \equiv 53 \pmod{97}$ 
So 
$\quad 11^{-1} \equiv 53 \pmod{97}$
But then
$\quad 2^{1}11^{-1} \equiv 9 \pmod{97}$
$\quad 2^{2}11^{-2} \equiv 81 \equiv -16 \equiv - 2^4 \pmod{97}$
and we can write down another relation,
$\tag 2 2^{-2}11^{-2} \equiv -1 \pmod{97}$
The two (primitive) relations are all we need.
$\quad \displaystyle {2^{11}}^{35} \equiv 11^{35} \equiv 11^{35} (-1) (11^{-34}) (2^{-34}) \equiv (-1) 11 \,{(2^{11})}^{-3}\, 2^{-1} \equiv (-1) \,({11}^{-2}) \, (2^{-1}) \equiv$
$\quad \quad (-1) \,\big [ ({11}^{-2}) \, (2^{-2}) \big ] \,2 \equiv (-1) \,(-1) \,2  \equiv 2 \pmod{97}$
