Scratch

Denote the function that is listed as $f (x)$ .

| ψ_{0} ⟩ = | 0 ⟩ | 1 ⟩

| ψ_{1} ⟩ = H | 0 ⟩ \otimes H | 1 ⟩ = (\frac{1}{\sqrt{2}} (| 0 ⟩ + | 1 ⟩)) \otimes (\frac{1}{\sqrt{2}} (| 0 ⟩ - | 1 ⟩)) = \frac{1}{2} (| 00 ⟩ - | 01 ⟩ + | 10 ⟩ - | 11 ⟩)

For the oracle stage, we pass through $| x ⟩ = \frac{1}{\sqrt{2}} (| 0 ⟩ + | 1 ⟩)$ . For $| y ⟩ = \frac{1}{\sqrt{2}} (| 0 ⟩ - | 1 ⟩)$ it becomes:

\begin{aligned} | y \oplus f (x) ⟩ & = \frac{1}{\sqrt{2}} (| 0 ⟩ - | 1 ⟩) \oplus | f (x) ⟩ \\ = \frac{1}{\sqrt{2}} (| 0 \oplus f (x) ⟩ - | 1 \oplus f (x) ⟩) \\ = \frac{1}{\sqrt{2}} (| f (x) ⟩ - | 1 \oplus f (x) ⟩) \\ = \frac{1}{\sqrt{2}} (| 1 ⟩ - | 0 ⟩) \\ = \frac{1}{\sqrt{2}} (- | 0 ⟩ + | 1 ⟩) \end{aligned}

| ψ_{2} ⟩ = \frac{1}{\sqrt{2}} (| 0 ⟩ + | 1 ⟩) \otimes \frac{1}{\sqrt{2}} (- | 0 ⟩ + | 1 ⟩) = \frac{1}{2} (- | 00 ⟩ + | 01 ⟩ - | 10 ⟩ + | 11 ⟩)

Last step:

| ψ_{3} ⟩ = H \frac{1}{\sqrt{2}} (| 0 ⟩ + | 1 ⟩) \otimes \frac{1}{\sqrt{2}} (- | 0 ⟩ + | 1 ⟩) = | 0 ⟩ \otimes \frac{1}{\sqrt{2}} (- | 0 ⟩ + | 1 ⟩) = \frac{1}{\sqrt{2}} (- | 00 ⟩ + | 01 ⟩)

Then measure and you always get $| 0 ⟩$ thus the function is a constant function.