= Solution
Let $V=\mathbb C^3$ be the defining <SU(3)> representation. A <mixed SU(3) tensor representation> belongs to $V^{\otimes k}\otimes(V^*)^{\otimes l}$, with transformation law
$$
T'^{\alpha_1\ldots\alpha_k}_{\beta_1\ldots\beta_l}
=U^{\alpha_1}{}_{\gamma_1}\cdots U^{\alpha_k}{}_{\gamma_k}
(U^{-1})^{\delta_1}{}_{\beta_1}\cdots(U^{-1})^{\delta_l}{}_{\beta_l}
T^{\gamma_1\ldots\gamma_k}_{\delta_1\ldots\delta_l}.
$$
The unrestricted <tensor> has <dimension> $3^{k+l}$. The upper and lower counts are independent; setting them equal would lose part of the printed generality.
For an irreducible <tensor>, impose the appropriate <Young symmetrizer> on each set of indices and subtract all upper-lower contractions. Antisymmetric pairs can be converted using the invariant alternating <tensor>, since $\bigwedge^2\mathbf3\cong\overline{\mathbf3}$ and $\bigwedge^2\overline{\mathbf3}\cong\mathbf3$. A full alternating triple is a singlet because $\det U=1$. <Traces> use the invariant <Kronecker delta> and yield lower-rank <tensors>. Repeating these operations gives invariant irreducible subspaces, with multiplicities retained.
An explicit algorithm for the complete decomposition starts at $(p,q)=(0,0)$, <tensors> with $\mathbf3$ exactly $k$ times, and then with $\overline{\mathbf3}$ exactly $l$ times. At each step apply
$$
(p,q)\otimes(1,0)=(p+1,q)\oplus(p-1,q+1)\oplus(p,q-1),
$$
$$
(p,q)\otimes(0,1)=(p,q+1)\oplus(p+1,q-1)\oplus(p-1,q),
$$
omitting terms with negative labels and adding multiplicities of coincident terms. The first is the <sl3 highest-weight tensor rule>: a Young diagram of row lengths $(p+q,q,0)$ gains a box in one of its three rows. The third-row term loses a <determinant> column of height three, leaving labels $(p,q-1)$. The second rule follows by conjugation, which exchanges $p$ and $q$. Thus the procedure determines all constituents of the general <tensor>, not just one selected symmetry type.
The separately symmetric, traceless component has labels $(k,l)$. Before imposing <traces> its <dimension> is $\binom{k+2}{2}\binom{l+2}{2}$. The contraction map has target the separately <symmetric tensors> of degrees $(k-1,l-1)$ and is onto: in polynomial coordinates it is $\sum_i\partial_{x_i}\partial_{y_i}$, the adjoint of multiplication by $\sum_ix_iy_i$ in the monomial-factorial <inner product>. Multiplication is injective, so the adjoint is surjective. Hence the <symmetric traceless SU(3) tensor representation> has <dimension>
$$
\boxed{d(k,l)=\binom{k+2}{2}\binom{l+2}{2}-\binom{k+1}{2}\binom{l+1}{2}
=\frac{(k+1)(l+1)(k+l+2)}2.}
$$
For $k=0$ or $l=0$ the subtracted term is zero. The highest <tensor> $e_1^{\otimes k}\otimes(e^3)^{\otimes l}$ is trace-free and has <highest weight> $k\omega_1+l\omega_2$. The three positive-root factors in the <Weyl dimension formula> are $k+1$, $l+1$ and $(k+l+2)/2$, giving exactly the kernel <dimension> above. Thus the whole symmetric traceless kernel is that irreducible constituent.
There is a genuine distinction between this highest-weight constituent and the literal largest-dimensional constituent. For example, $\mathbf3^{\otimes5}$ contains $(3,1)$, obtained by antisymmetrizing one pair, with <dimension> $24$, whereas the fully symmetric $(5,0)$ has <dimension> $21$. Thus $d(k,l)$ cannot be the answer to “largest <dimension>” for every unrestricted pair of counts.
The <maximal dimension in mixed SU(3) tensor powers> can nevertheless be determined exactly. Every constituent <highest weight> differs from $k\omega_1+l\omega_2$ by $a\alpha_1+b\alpha_2$ with nonnegative integers $a,b$, where $\alpha_1=2\omega_1-\omega_2$ and $\alpha_2=-\omega_1+2\omega_2$. Its labels therefore satisfy
$$
p=k-2a+b\ge0,\qquad q=l+a-2b\ge0.
$$
If both $a,b$ are positive, subtracting $c=\min(a,b)$ from both increases the labels to $(p+c,q+c)$, whose <dimension> is strictly larger. The resulting boundary labels occur in the <tensor>: use $a$ disjoint antisymmetric upper pairs or $b$ disjoint antisymmetric lower pairs, and then take the separately symmetric traceless component. Consequently the exact answer is the finite formula
$$
\boxed{D_{\max}(k,l)=\max\left\{
\max_{0\le a\le\lfloor k/2\rfloor}d(k-2a,l+a),\quad
\max_{0\le b\le\lfloor l/2\rfloor}d(k+b,l-2b)\right\}.}
$$
This supplies a practical closed finite computation for arbitrary $k,l$. To reduce it to at most four evaluations, put $K=k+1$, $L=l+1$ and $r=1+\sqrt3$. The first continuous maximum is at $a_*=(K-rL)/(2+r)$, clipped to $[0,k/2]$; test its two neighbouring allowed integers. The second is similarly at $b_*=(L-rK)/(2+r)$, clipped to $[0,l/2]$. Indeed differentiating the first cubic gives a derivative with the sign of $(K-2a)^2-2(K-2a)(L+a)-2(L+a)^2$, which changes sign once at that root. For equal upper and lower counts, both maxima are at zero and the familiar result is $(k+1)^3$.
For three <quarks>, the unconstrained flavour product is
$$
\mathbf3\otimes\mathbf3\otimes\mathbf3
=(\mathbf6\oplus\overline{\mathbf3})\otimes\mathbf3
=\mathbf{10}_{[3]}\oplus\mathbf8_{[21]}\oplus\mathbf8_{[21]}\oplus\mathbf1_{[111]}.
$$
The brackets indicate permutation symmetry: symmetric, mixed, and alternating. The <three-quark colour singlet> is proportional to $\epsilon_{abc}$ and is alternating. The <fermion> wavefunction must be alternating overall. In the usual spatially symmetric ground-state S-wave, the spin-flavour part must therefore be symmetric.
The three-spin-one-half space has a symmetric spin-$3/2$ sector and mixed-symmetry spin-$1/2$ sectors; it has no fully alternating sector because $\bigwedge^3\mathbb C^2=0$. Symmetric flavour can pair with symmetric <spin>, giving the <baryon decuplet> of <spin> $3/2$. The flavour-octet multiplicity space and the spin-$1/2$ multiplicity space both transform as the two-dimensional standard representation of the <symmetric group>. Their product contains exactly one symmetric combination: its characters at the identity, a transposition and a three-cycle are $4,0,1$, so the symmetric multiplicity is $(4+3\cdot0+2\cdot1)/6=1$. This gives one <baryon octet> of <spin> $1/2$. Alternating singlet flavour would need alternating <spin> to make the spin-flavour part symmetric, and that sector does not exist. Therefore the <Pauli constraint on three-quark flavour multiplets> yields
$$
\boxed{\mathbf8\text{ with }J^P=\tfrac12^+,\qquad\mathbf{10}\text{ with }J^P=\tfrac32^+,\qquad\text{no flavour singlet in the symmetric spatial ground state}.}
$$
The spin-flavour <dimension> check is $8\cdot2+10\cdot4=56=\dim\operatorname{Sym}^3\mathbb C^6$, the <symmetric spin-flavour SU6 representation>.
The spatial-symmetry qualification matters if the wording is interpreted as total $L=0$ alone. Rotationally scalar excited orbital functions can have mixed permutation symmetry. For example, the two independent differences among $r_{12}^2,r_{23}^2,r_{31}^2$, multiplied by a symmetric radial factor, span such a space while remaining rotational scalars. Combining this standard permutation representation with the spin-$1/2$ standard representation gives an alternating spin-orbital combination; multiplying by alternating singlet flavour makes it symmetric, and the colour factor restores overall fermionic antisymmetry. Thus total $L=0$ without a symmetric ground-state orbital assumption would not by itself exclude a singlet.
For a quark-antiquark pair, colour singletness selects the contraction $\delta^a{}_b$, and the independent flavour product gives
$$
\boxed{\mathbf3_F\otimes\overline{\mathbf3}_F=\mathbf8_F\oplus\mathbf1_F,\qquad
\tfrac12\otimes\tfrac12=0\oplus1.}
$$
<Quark> and <antiquark> are distinguishable, so there is no three-identical-quark exclusion of either flavour multiplet or either <spin>. Both the octet and singlet admit <spin> singlets and <spin> triplets. Their <baryon number> is zero. The flavour octet has rows $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$, while the singlet has $(Y,I)=(0,0)$. The octet's zero-isospin state is $(u\bar u+d\bar d-2s\bar s)/\sqrt6$, and the singlet is $(u\bar u+d\bar d+s\bar s)/\sqrt3$. Each row contains all $I_3=-I,\ldots,I$, and <electric charge> is $Q=I_3+Y/2$.
For ground-state <mesons> with $L=0$, opposite intrinsic quark-antiquark parities give $P=-1$. The neutral self-conjugate states have <charge conjugation> $C=(-1)^{L+S}$, so \b[both flavour multiplets occur as $0^{-+}$ <pseudoscalar mesons> and $1^{--}$ <vector mesons>], forming a flavour nonet in each channel. Charged and open-flavour states have no individual charge-conjugation <eigenvalue>. The <meson> request does not restrict orbital excitation: in general $J=|L-S|,\ldots,L+S$, $P=(-1)^{L+1}$, and the same formula for $C$ applies where defined. Thus the flavour representations stay $\mathbf1\oplus\mathbf8$, with these additional orbital-spin quantum numbers.
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