 Solution

Choose a perfect matching $M$ from (i). Every vertex has exactly two incident edges in $E\setminus M$, so this spanning subgraph is a disjoint union of graph cycles. Orient each graph cycle cyclically, and let $s(v)$ be the next vertex after $v$ along its oriented graph cycle.

For each $uv\in M$, take the three-edge sequenceIts four vertices are distinct. Each $s(u)\ne u$ and $s(v)\ne v$ because the graph is a simple graph. Also $s(u)\ne v$ and $s(v)\ne u$, since $uv$ belongs to $M$, rather than $E\setminus M$. Finally $s(u)\ne s(v)$ because the successor map $s$ is a bijection on the disjoint graph cycles. Thus this is a graph path of length exactly three.

Every edge of $M$ is used once as a middle edge. Every edge of $E\setminus M$ is used once as the outgoing edge of its tail, whose unique edge of $M$ determines which graph path contains it. Hence these graph paths are pairwise edge-disjoint and cover $E$. This is the three-edge path decomposition from a perfect matching.