= Solution
The root-of-unity specialization requires $q\ne1$: we take $q$ to be a primitive cube <root of unity>. The displayed presentation contains $(q-q^{-1})^{-1}$, so substituting $q=1$ literally is undefined. Its classical $U(\mathfrak{sl}_2)$ limit instead has simple finite-dimensional <modules> of arbitrarily large dimension. This is why the primitive-root qualification matters.
First consider <simple modules> occurring in the <quantum plane>. Here $[3]=0$, $K^3=1$, and
$$
E^3(x^ry^s)=[s][s-1][s-2]x^{r+3}y^{s-3}=0,
$$
with the analogous formula $F^3=0$. Thus the action factors through the finite-dimensional quotient with $E^3=F^3=0$, $K^3=1$, spanned by $F^iK^jE^l$ with $0\le i,j,l<3$. Any <simple module> over this quotient is finite-dimensional, since it is generated by one vector. In a nonzero simple <submodule> choose a $K$ <eigenvector> $v$ in $\ker E$. This is possible because $E$ is nilpotent and its kernel is $K$-invariant. Then the <highest-weight string for quantum sl2> shows that
$$
W=\langle v,Fv,F^2v\rangle
$$
is invariant under $E,K$, and $F^3v=0$ makes it $F$-invariant. Simplicity gives $W=V$, proving \b[$\dim V\le3$].
The same bound holds even for arbitrary finite-dimensional <simple modules> over the unrestricted <quantum enveloping algebra of sl2>. Indeed $E^3,F^3,K^3$ are central: the string commutation identity has a factor $[3]=0$, and the $K$ commutation factors satisfy $q^6=1$. By <Schur lemma>, they act as scalars $\alpha,\beta,\kappa$, with $\kappa\ne0$. The element
$$
\Omega=FE+\phi(K),\qquad \phi(K)=\frac{qK+q^{-1}K^{-1}}{(q-q^{-1})^2}
$$
is also central. For example $[E,FE]=(K-K^{-1})E/(q-q^{-1})$, while $[E,\phi(K)]=[\phi(q^{-2}K)-\phi(K)]E=-(K-K^{-1})E/(q-q^{-1})$; the $F$ commutator cancels in the same way. Let $\Omega$ act as $\gamma$.
If $\alpha=0$, choose the highest vector $v$ as above. Its three-vector $F$ string is invariant, now using $F^3v=\beta v$, so again its dimension is at most three. If $\alpha\ne0$, choose a $K$ <eigenvector> $v$. The space $W=\langle v,Ev,E^2v\rangle$ is stable under $E,K$ because $E^3=\alpha$, and it is stable under $F$ because
$$
F=(\gamma-\phi(K))E^{-1},\qquad E^{-1}=\alpha^{-1}E^2.
$$
It therefore equals the simple <module> and has dimension at most three.
Over $\mathbb C$ this also excludes infinite-dimensional <simple modules> of the unrestricted algebra. It is a finite <module> over the central finitely generated commutative <algebra over a field> $Z_0=\mathbb C[E^3,F^3,K^3,K^{-3}]$, using the same 27 ordered spanning elements. A simple <module> $M$, generated by one vector, is consequently finite over $Z_0$. Choose a <maximal ideal> $\mathfrak m$ in its support. <Nakayama lemma> makes $M/\mathfrak mM\ne0$; since $\mathfrak mM$ is a <submodule>, simplicity gives $\mathfrak mM=0$. The <Hilbert Nullstellensatz> gives $Z_0/\mathfrak m=\mathbb C$, so $M$ is finite-dimensional and the preceding argument applies. Thus the usual complex, primitive-root interpretation yields
$$
\boxed{\text{every simple }U_q\text{-module has dimension at most }3.}
$$
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