Both sides are antisymmetric in $(i,j)$ and in $(p,q)$, and both vanish unless $\{i,j\} = \{p,q\}$ (as subsets). It suffices to check the two cases: **Case $i = p$, $j = q$ with $i \neq j$:** The right side gives $\delta_{ii}\delta_{jj} - \delta_{ij}\delta_{ji} = 1 \cdot 1 - 0 \cdot 0 = 1$. The left side: $\varepsilon_{ijk}\varepsilon_{ijk} = \sum_k \varepsilon_{ijk}^2$. Since $i \neq j$, there is exactly one value of $k$ (namely the third index not equal to $i$ or $j$) for which $\varepsilon_{ijk} \neq 0$, and $\varepsilon_{ijk}^2 = 1$. So the left side equals $1$. $\checkmark$ **Case $i = q$, $j = p$ with $i \neq j$:** The right side gives $\delta_{ij}\delta_{ji} - \delta_{ii}\delta_{jj} = 0 - 1 = -1$. The left side: $\varepsilon_{ijk}\varepsilon_{jik} = -\varepsilon_{ijk}\varepsilon_{ijk} = -1$. $\checkmark$