= Solution
{c}
Let $a_{ij}$ be the coefficient of $|ij\rangle$. The <partial trace> contracts the repeated index of the discarded <qubit>:
$$
(\rho_1)_{ii'}=\sum_{j=0}^1a_{ij}\overline{a_{i'j}},
\qquad
(\rho_2)_{jj'}=\sum_{i=0}^1a_{ij}\overline{a_{ij'}}.
$$
For the first <qubit>, each diagonal entry is $1/3+1/6=1/2$ and the upper off-diagonal entry is $1/3-1/6=1/6$. For the second <qubit>, the diagonal entries are $2/3$ and $1/3$, while the upper off-diagonal entry is $-1/\sqrt{18}+1/\sqrt{18}=0$. Thus the two <reduced density matrices>, each in <computational basis> order $0,1$, are
$$
\boxed{\rho_1=\operatorname{Tr}_2\rho=\begin{pmatrix}1/2&1/6\\1/6&1/2\end{pmatrix},\qquad
\rho_2=\operatorname{Tr}_1\rho=\begin{pmatrix}2/3&0\\0&1/3\end{pmatrix}.}
$$
Both have <eigenvalues> $2/3,1/3$ and <trace> one, so both are <mixed states> even though the joint <quantum state> is pure. Their different <eigenvectors> are visible in the useful rewrite
$$
|\psi\rangle=\sqrt{\frac23}|+\rangle|0\rangle-\sqrt{\frac13}|-\rangle|1\rangle,
\qquad |\pm\rangle=\frac{|0\rangle\pm|1\rangle}{\sqrt2}.
$$
Absorbing the minus sign into one local basis vector makes this a <Schmidt decomposition> with nonnegative coefficients. It also checks both <partial traces> and exhibits the <entanglement> responsible for the mixed local descriptions.
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