It looks like you missed the promising reformulation of this 3SAT. The simple idea is to select a different variable for each clause.
Always satisfiable
Let the variables be $V=\{v_1, v_2, \cdots, v_n\}$ and the clauses be $C=\{c_1,c_2,\cdots, c_m\}$.
Construct a bipartite graph $G=(C,V)$, where $(c,v)$ is an edge if $v$ or $\neg v$ is a literal of $c$. The given conditions mean the degree of each clause and the degree of each variable are 3.
For a subset $W$ of $C$, let $N(W)$ denote the set of all variables adjacent to some clause in $W$. Consider the edges that have one endpoint in $W$. The number of them is exactly $3|W|$ and at most $3|N(W)|$. Hence $|W|\leq |N(W)|.$ By Hall's marriage theorem, for each clause $c$, we can select a distinct variable $m(c)$ such that $(c,m(c))$ is an edge.
For all clause $c$ do the following. If $m(c)$ is a literal of $c$, set $m(c)$ to be true. Otherwise, $\neg m(c)$ is a literal $c$ and we set $m(c)$ to be false. In either case, $c$ becomes true.
Polynomial algorithm
The proof above actually gives the outline of an algorithm. Each step of the algorithm takes polynomial time. In particular, the application of Hall's marriage theorem can be implemented in polynomial time as, for example shown here.
A generalization
Here is a natural generalization.
Exercise. Generalize to the case when each variable appears in at most $r$ clauses and each clause contains at least $r$ distinct variables, where $r\ge1$.
