Let $G=(V,E)$ be a graph on $2n$ vertices i.e. $|V(G)|=2n$. Moreover, the graph $G$ is assumed to be a $3$-partite graph such that $V(G)=\bigcup_{i=1}^{3}V_i$ where $V_i$ denotes the $i$th partite set. We also assume that $|V_1|\ge |V_2|\ge |V_3|$ and that $|V_1|<n$.
We then take a partition $V(G)=W_1\cup W_2$ such that $V_1\subseteq W_1\subseteq V_1\cup V_3$ and also $V_2\subseteq W_2\subseteq V_2\cup V_3$. Now, let $\mathcal{E}(W_k)$ for $k\in\{1,2\}$ be the number of edges such that both endpoints of each edge in $\mathcal{E}(W_k)$ is in $W_k$ for $k\in\{1,2\}$.
It is stated that $\mathcal{E}(W_1)\le |V_1|\cdot (n-|V_1|)$ (and similarly that $\mathcal{E}(W_2)\le |V_2|\cdot(n-|V_2|)$ assuming that the partition $V(G)=W_1\cup W_2$ is a balanced partition (i.e. a partition such that the cardinalities of $W_1$ and $W_2$ are as close to each other as possible). However, I am not exactly sure where this follows from, even though I've tried to think about it. My idea simply is that because of the assumption that $|V_1|$ is the largest of the cardinalities of the partite sets, it follows that the cardinality of $V_1$ multiplied by the quantity $(n-|V_1|)$ is clearly at least $|V_1|$. However, this is confusing because as stated earlier, we have $V_1\subseteq W_1\subseteq V_1\cup V_3$ (which is clearly used to derive the fact that $\mathcal{E}(W_1)\le |V_1|\cdot (n-|V_1|)$).
I think this statement is incorrect. In fact, consider the graph $G=(V,E)$ where $V=V_1\cup V_2\cup V_3$ and $|V_1|=|V_2|=|V_3|=2n/3$. The set of edges $$ E=\{(x,y)\mid x\in V_i,\ y\in V_j,\ i\neq j,\ i=1,2,3;j=1,2,3\}. $$ Let $W_1=V_1\cup V_3$, $W_2=V_2$. We have $E(W_1)=|V_1|\cdot|V_3|=4n^2/9$, but $|V_1|(n-|V_1|)=2n^2/9$. Hence $E(W_1)>|V_1|(n-|V_1|)$.
Addition.
If a partition such that the cardinalities of $W_1$ and $W_2$ are as close to each other as possible, then $W_1=V_1\cup V'_3$ and $W_2=V_2\cup V''_3$, where $V_3=V'_3\cup V''_3$, $V'_3\cap V''_3=\varnothing$ and $|V_1|+|V'_3|\leq n$ (otherwise you can decrease $W_1$ and increase $W_2$, so that $||W_1|-|W_2||$ will decrease). Then it is clear that $$ E(W_1)\leq |V_1|\cdot|V'_3|\leq|V_1|(n-|V_1|). $$