Levi Cevita and Kronecker Delta identity
I think it's helpful to see how you can actually derive this identity, using a different definition of $\epsilon_{ijk}$. I hope you are a friend of matrices and determinants, since I am going to use that a lot in what follows now. (I will not be using the Einstein summing convention in the proof, but then again in the other parts of this answer) $\newcommand{\vek}[1]{\boldsymbol{#1}}$ The most common definition of $\epsilon_{ijk}$ is $$ \epsilon_{ijk} = \begin{cases} 1 & \text{if $\space (ijk) \space$ is an even permutation of (123) } \\ -1 & \text{if $\space (ijk) \space$ is an odd permutation of (123) } \\ 0 & \text{else} \end{cases} $$ Simply comparing the following two expressions, you can observe that
$$ \det (\vek{e}_i ~\vek{e}_j ~\vek{e}_k) = \epsilon_{ijk} ~~~~\text{for all}~~~ (ijk) $$
Armed with this, and some linear algebra, lets tackle the identity you are interested in:
$$ \sum_{i=1}^3 \epsilon_{ijk}\epsilon_{ilm} = \delta_{lj} \delta_{mk} - \delta_{lk} \delta_{mj} $$
Proof
$\color{blue}{\text{If } j=k}$ then no matter the value of $i$ , every summand will be $0$ and thus the sum as a whole. So we have $$ \delta_{lj} \delta_{mk} - \delta_{lk} \delta_{mj} = \delta_{lk} \delta_{mk} - \delta_{lk} \delta_{mk} = 0 = \sum_{i=1}^3 \epsilon_{ijk}\epsilon_{ilm} $$
$\color{blue}{\text{If } j \neq k}$ then there is one value of $i \in \{1,2,3\}$ , such that $i,j,k$ are all different. For this value of $i$, we have
\begin{align*} \epsilon_{ijk}\epsilon_{ilm} &= \det( \vek{e}_i ~\vek{e}_j ~\vek{e}_k) \det \begin{pmatrix} \vek{e}_i^T \\ \vek{e}_l^T \\ \vek{e}_m^T \\ \end{pmatrix} = \det \begin{pmatrix} \vek{e}_i^T \vek{e}_i & \vek{e}_i^T \vek{e}_j & \vek{e}_i^T \vek{e}_k \\ \vek{e}_l^T \vek{e}_i & \vek{e}_l^T \vek{e}_j & \vek{e}_l^T \vek{e}_k \\ \vek{e}_m^T \vek{e}_i & \vek{e}_m^T \vek{e}_j & \vek{e}_m^T \vek{e}_k \\ \end{pmatrix} \\&= \det \begin{pmatrix} \delta_{ii} & \delta_{ij} & \delta_{ik} \\ \delta_{li} & \delta_{lj} & \delta_{lk} \\ \delta_{mi} & \delta_{mj} & \delta_{mk} \\ \end{pmatrix} = \det \begin{pmatrix} 1 & 0 & 0 \\ \delta_{li} & \delta_{lj} & \delta_{lk} \\ \delta_{mi} & \delta_{mj} & \delta_{mk} \\ \end{pmatrix} \\&= \det \begin{pmatrix} \delta_{lj} & \delta_{lk} \\ \delta_{mj} & \delta_{mk} \\ \end{pmatrix} = \delta_{lj} \delta_{mk} - \delta_{lk} \delta_{mj} \end{align*} For the other values of $i$, we either have $i=j$ or $i = k$ and thus $\epsilon_{ijk}=0$ . Thus, summing over all values of $i$, we get $$ \sum_{i=1}^3 \epsilon_{ijk}\epsilon_{ilm} = \delta_{lj} \delta_{mk} - \delta_{lk} \delta_{mj} $$
Which completes the proof.
Mnemonic
\begin{align} \require{cancel} \epsilon_{ijk}\epsilon_{ilm} &= \det(\vek{e}_i ~\vek{e}_j ~\vek{e}_k) ~\det(\vek{e}_i ~\vek{e}_l ~\vek{e}_m) \\ &= \color{blue}{\det(\cancel{\vek{e}_i} ~\vek{e}_j ~\vek{e}_k) ~\det(\cancel{\vek{e}_i} ~\vek{e}_l ~\vek{e}_m) = \det(\vek{e}_j ~\vek{e}_k) ~\det(\vek{e}_l ~\vek{e}_m) } \\ &= \det(\vek{e}_j ~\vek{e}_k) ~\det \begin{pmatrix} \vek{e}_l^T \\ \vek{e}_m^T \end{pmatrix} \\ &= \det \begin{pmatrix} \vek{e}_l^T \vek{e}_j & \vek{e}_l^T \vek{e}_k \\ \vek{e}_m^T \vek{e}_j & \vek{e}_m^T \vek{e}_k \\ \end{pmatrix} \\ &= \det \begin{pmatrix} \delta_{lj} & \delta_{lk} \\ \delta_{mj} & \delta_{mk} \\ \end{pmatrix} = \delta_{lj} \delta_{mk} - \delta_{lk} \delta_{mj} \end{align} You could just remember what is marked in blue; the $\color{blue}{\text{formal canceling of } \vek{e}_i}$ . The calculation after that is pretty straight forward, and you get the correct result.
Usage of the identity
Where would you use this identity? For example if you want to prove $$ \vek{a} \times (\vek{b} \times \vek{c} ) = \vek{b} (\vek{a} \cdot \vek{c} ) - \vek{c} ( \vek{a} \cdot \vek{b} ) $$ Because you can start at the left side of the equation (here, using Einstein summation!) \begin{align} \vek{a} \times (\vek{b} \times \vek{c}) &= \epsilon_{ijk} a_k (\vek{b} \times \vek{c})_i ~\vek{e}_j \\ &= \epsilon_{ijk} a_k \left( \epsilon_{ilm} b_l c_m \right) ~\vek{e}_j \\ &= (\epsilon_{ijk}\epsilon_{ilm}) ~a_k b_l c_m ~\vek{e}_j \end{align} where you can now use the identity to continue.
Here is a new reference
Another reference explaining how one can generalize Levi Civita symbol.