 Solution

Taking partial inner products with $⟨0|$ and $⟨1|$ on the first wire gives $(|0⟩+|1⟩)/2$ and $(-|0⟩+|1⟩)/2$. They are nonzero and orthogonal, hence linearly independent. This contradicts the necessary product-state property proved in part (a), so the second state is entangled.

Equivalently, its tensor-product coefficient matrix is $C=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$, with $\det C=1/2\ne 0$. The two-qubit product-state determinant criterion gives the same conclusion. In fact $C{C}^{†}=I/2$, so its Schmidt coefficients are both $1/\sqrt{2}$: it is maximally entangled.