Is every X3SAT instance with no cycles satisfiable?
The graph below is a positive answer without words.


Here is the detailed proof.
Definitions
Let $X$ be an instance of X3SAT.

$X$ is linear if any two clause shares at most one variable.
$X$ is monotone if all literals in all clauses are positive.
A walk in $X$ is a sequence of clauses where each clause is not disjoint with the clause following it. The length of the walk is the number of its clauses. A walk starts with variable $v$ if its first clause contains $v$. A walk ends with variable $v$ if its last clause contains $v$. A walk is said to connect two variables if it starts with one of the variables and ends with the other variable. In particular, any clause connects all variables in it.
A trail is a walk in which all clauses are distinct.
$X$ is connected if there is a walk between any two variables.
A cycle is a walk whose first clause is its last clause. A simple cycle has no repeated clauses except the first and last clause. $X$ is acyclic if $X$ has no simple cycles, i.e., there is at most one trail between any two variables,
For any two variables, the distance between them is the minimum length of a walk that connects them. The distance between a variable and itself is 0.

A linear monotone acyclic instance of X3SAT is always satisfiable.
Let $X$ be a linear monotone acyclic instance of X3SAT whose set of variables is $V=\{v_1,v_2, \cdots, v_n\}$. Because of monotonicity, we will use variable and literal interchangeably. Each clause in $X$ is also treated as the set of three variables in it.
Assume $X$ is connected. Otherwise, $X$ is the union of two disjoint linear monotone acyclic instances of X3SAT that can be treated independently.
Let $d(u)$ denote the distance from $v_1$ to variable $u$.
Claim 1. For any clause $c=(t,u, w)$, $\{d(t), d(u), d(w)\}=\{\ell, \ell, \ell-1\}$ for some $\ell\ge1$.
Proof: If $c$ contains $v_1$, then it is easy to see that $\{d(t), d(u), d(w)\}=\{1, 1, 0\}$. Now assume $c$ does not contain $v_1$.
WLOG, let $d(w)=\ell-1 &gt; 0$ be the minimum among the three distances. Let $c_1, \cdots, c_{\ell-2}, c_{\ell-1}$ be the unique trail from $v_1$ to $w$. The shared variable between $c_{\ell-2}$ and $c_{\ell-1}$ cannot be $t$ or $u$, since $d(t)\ge \ell-1$ and $d(u)\ge\ell-1.$ That means $c_{\ell-1}\not=c$. Suppose there is a trail $b_1, b_2, \cdots, b_{\ell-1}$ from $v_1$ to $t$. Then for the same reason, $b_{\ell-1}\not=c$. We can reduce the walk $c_1, \cdots, c_{\ell-2}, c_{\ell-1}, c, b_{\ell-1}, b_{\ell-2}, \cdots, b_1, c_1$ to a simple cycle, which is a contradiction, which implies that there is no trails of length $\ell-1$ from $v_1$ to $t$, i.e., $d(t)\gt\ell-1$. Since $c_1, c_2, \cdots, c_{\ell-1}, c$ is a walk of length $\ell$ from $v_1$ to $t$,  $d(t)=\ell$. Similarly, $d(u)=\ell.$ QED.
Fix a total order $\prec$ on $V$. For example, we can let $v_p\prec v_q $ if $p&lt;q$.
Let $A_0=\{v_0\}$. For all $\ell\ge 1$, define $A_{\ell}$ recursively by
$$A_\ell=
\{t\in V: d(t)=\ell \text{ and the last clause of the shortest trail}\\
\text{ from } v_1\text{ to }t\text{ is }(t,u, w)\text{ for some }u,w\in V \text{ such that }\\
d(t)=d(u)=\ell\text{ and }d(w)=\ell-1\text{ and } w\not\in A_{\ell-1}\text{ and }t\prec u\}$$
Let $A=\cup_{\ell=0}^{\infty}A_\ell$. Note that if $d(t)=\ell$, then $t\in A$ iff $t\in A_\ell$.
Claim 2: $A$ shares exactly one variable with each clause.
Proof. Let $c$ be a clause in $X$. By claim 1, $c=(t,u,w)$ where $d(t)=d(u)=\ell$, $d(w)=\ell-1$ and $t\prec u$. The last clause of the shortest trail from $v_1$ to either $t$ or $u$ is $c$.

$w\in A_{\ell-1}$. By definition, $t$ and $u$ are not in $A_\ell$.
$w\not\in A_{\ell-1}$. By definition, $t\in A_\ell$ and $u\not\in A_\ell$.

QED.

Exercises
Let X be a linear monotone acyclic instance of X3SAT with $n$ variables and $m$ clauses.
Exercise 1..(One minute or two) Explain that $A_1$ is empty. Give an example where $A_3$ is not empty.
Exercise 2.. Show that $n= 2m+1$.