Polynomial-time linear-reduction from Directed Hamiltonian Path Problem to 3SAT
I don't know of a reference, but there's a standard reduction from Hamiltonian paths/cycles to SAT which is fairly well-known. The reduction from SAT to 3SAT is then done by introducing additional variables to break up clauses, e.g.:

$$v_1 \vee \neg v_2 \vee v_3 \vee \neg v_4 \Rightarrow \left( v_1 \vee \neg v_2 \vee X\right) \wedge \left( v_3 \vee \neg v_4 \vee  \neg X\right)$$

Let $n$ be the number of nodes in the graph. Then define variables $x_{i,j}$ where $i,j \in \left\{1 \ldots n\right\}$, where $x_{i,j}$ is true if node $i$ appears in position $j$ in the Hamiltonian path.

Every node must appear somewhere on the path:

$$\bigwedge_{i} \left( \bigvee_{j} x_{i,j}\right)$$

Every position in the path must be occupied:

$$\bigwedge_{i} \left( \bigvee_{j} x_{j,i}\right)$$

No node appears on the path more than once:

$$\bigwedge_{i} \bigwedge_{j \ne k} \left( \neg x_{i,j} \vee \neg x_{i,k} \right)$$

No two distinct nodes occupy the same position:

$$\bigwedge_{i} \bigwedge_{j \ne k} \left( \neg x_{j,i} \vee \neg x_{k,i} \right)$$

Nonadjacent nodes cannot be adjacent on the path. If $E \subseteq N \times N$ is the set of edges:

$$\bigwedge_{k \ne \left|N\right|}\bigwedge_{(i,j) \not\in E} \left( \neg x_{k,i} \vee \neg x_{k+1,j}\right)$$

You can modify this in the obvious way if you're trying to find a Hamiltonian cycle.

On that, as a practical matter, if you are trying to find a Hamiltonian cycle and you intend to feed this to a SAT solver, you would normally assert a start node as another clause, e.g. $x_{1,1}$, to reduce the search space. Obviously for Hamiltonian paths, you don't do that.