The exercise in question is from Hunter's Applied Analysis, specifically exercise 5.2.
Here is given an n-dimensional space $X$, with bases $\{e_1,\ldots,e_n\}$ and $\{\tilde{e}_1,\ldots,\tilde{e}_n\}$. The bases are related by a matrix $(L_{ij})$ with inverse $(\tilde{L}_{ij})$, such that $$ \tilde{e}_i = \sum_{j=1}^n L_{ij}e_j;\qquad e_i = \sum_{j=1}^n \tilde{L}_{ij}\tilde{e}_j;\qquad \delta_{ik} = \sum_{j=1}^n L_{ij}\tilde{L}_{jk}, $$ where $\delta_{ik}$ is the Kronecker delta function.
Additionally we are given the dual bases $\{\omega_1,\ldots,\omega_n\}$ and $\{\tilde{\omega}_1,\ldots,\tilde{\omega}_n\}$ of $X^*$.
To prove:
a) If $x=\sum x_ie_i = \sum\tilde{x}_i\tilde{e}_i\in X$, then $$\tilde{x}_i = \tilde{L}_{ij}x_j.$$
b) If $\phi=\sum\phi_i\omega_i = \sum\tilde{\phi}_i\tilde{\omega}_i\in X^*$, then $$\tilde{\phi}_i=L_{ji}\phi_j.$$
I am still stuck at a) (and would not appreciate answers for b) just yet, maybe perhaps small hints), and what I managed to derive is that $$ \sum_{i=1}^nx_ie_i = \sum_{i=1}^nx_i\sum_{j=1}^n\tilde{L}_{ij}\tilde{e}_j = \sum_{j=1}^n\tilde{e}_j\sum_{i=1}^n\tilde{L}_{ij}x_i=\sum_{j=1}^n\tilde{e}_j\tilde{x}_j, $$ which implies that $\tilde{x}_i = \sum_{j=1}^n\tilde{L}_{ji}x_j$. This should make sense, and every textbook and online source I searched lead me to a similar answer, which contradicts a). Notice that even the indices are switched, and nothing is stated about $j$ in the question. Am I missing something here, or there is simply an error in the exercise?
I don't think you're missing anything here, and I think Kavi is right...or the book made an error.
In particular, you can prove that the assumptions imply that $\overline{x}_j=L_{ij}x_j$ for all $j$ does not necessarily happen.
Your derivation shows that $$\sum_{j=1}^n \overline{x}_j\overline{e}_j=\sum_{j=1}^n \overline{e}_j\left[ \sum^n_{i=1} \overline{L}_{ij} x_i \right],$$ hence $\overline{x}_j=\sum^n_{i=1} \overline{L}_{ij} x_i$. So $\overline{x}_j=\overline{L}_{ij}x_j$, for all $j$ implies that $\overline{x}_j=0$, hence $x=0$.
But we have the simple counterexample of $X=\mathbb{R}$, $e_1=\overline{e}_1=1$, $L_{ij}=[1]$, and $x=1$, showing that this doesn't necessarily happen.
I think it's safe to assume what Kavi did and interpret the book as using the Einstein summation convention.
I'm going to respect your wishes and not give an answer (or a hint) for part (b).