= Solution
The <finite representation type of a quiver> condition means there are only finitely many isomorphism classes of indecomposable finite-dimensional $kQ$-modules. Equivalently, every finite-dimensional <module> is a <direct sum> of members of a finite list of indecomposables.
For the $E_6$ case, we give the reflection argument underlying the <Gabriel theorem>, rather than using its conclusion as the proof. Since the underlying graph is a tree, every orientation is acyclic. At a sink $i$, write
$$
h:\bigoplus_{j\to i}V_j\longrightarrow V_i
$$
for the sum of the incoming arrow maps. If an indecomposable $V$ is not the simple $S_i$, then $h$ is surjective: a nonzero complementary subspace to its image at the sink would be a <direct summand> supported only at $i$. The <Bernstein–Gelfand–Ponomarev reflection functor> reverses the arrows and replaces $V_i$ by $\ker h$, with reversed maps the component projections. The source reflection replaces that <kernel> by the <cokernel> of its inclusion into $\bigoplus V_j$, recovering $V_i$. The inclusion is injective, so the reflected representation has no <direct summand> $S_i$ at the new source. If it decomposed, applying the inverse would decompose $V$. Thus reflection preserves indecomposability except that it annihilates $S_i$.
On a <dimension vector of a quiver representation>, it acts by the integral simple reflection
$$
(s_i d)_i=-d_i+\sum_{j\sim i}d_j,\qquad (s_i d)_j=d_j\quad(j\ne i).
$$
Let $B=2I-\operatorname{Adj}(E_6)$. These reflections preserve the <positive-definite quadratic form> $d^{\mathsf T}Bd$. Positivity can be checked without a classification theorem: label the graph by the chain $1-2-3-4-5$ and attach vertex $6$ to $3$. The leading principal minors are $2,3,4,5,6,3$, all positive.
Choose a topological ordering so arrows point from smaller to larger indices. Reflect successively at sinks $6,5,\ldots,1$. This restores the orientation after one full cycle. The transformation of dimension vectors is the <Coxeter element> $C=s_1s_2\cdots s_6$. If $U$ has diagonal entries one and entry $-1$ above the diagonal at an arrow, direct multiplication of the reflection matrices gives
$$
C=-U^{-1}U^{\mathsf T},\qquad I-C=U^{-1}(U+U^{\mathsf T})=U^{-1}B.
$$
Thus $C$ has no fixed vector. It is an integral isometry of a positive definite form, so it has finite order $h$: the image of each lattice <basis vector> lies in a finite set of vectors of the same norm, giving only finitely many possible isometries. Since $C^h=I$ and $I-C$ is invertible,
$$
I+C+\cdots+C^{h-1}=0.
$$
If an indecomposable survived all $h$ reflection cycles, its dimension vectors $d,Cd,\ldots,C^{h-1}d$ would all be nonzero and componentwise nonnegative, contradicting their zero sum. Therefore it reaches a simple at one of the finitely many reflection steps and is then annihilated. Applying the inverse reflections uniquely reconstructs it from that simple. There are at most $6h$ such stopping positions, proving
$$
\boxed{kQ_3\text{ has finite representation type for every orientation of }E_6.}
$$
This supplies the requested finite-type proof sketch.
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