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Currently, I am reading Representation Theory of Semi-Direct Products by Reyes.

In section $6$, the author mentions that the presentation of $D_4 \rtimes Aut(D_4)$ is as follows.

$$ D_4 \rtimes Aut(D_4) = \left\{a, b, x, y | a^4 = b^2 = x^4 = y^2 = e, b a b = a^3, x^3 a x = a, x^3 b x = a b, y a y = a^3, y b y = a b \right\} $$

Here is the screenshot.

enter image description here

It is not obvious to me how the presentation is like this. Any help will be appreciated.

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    Could it be that the relation $yxy = x^3$ should be added? I am unable to reach the original article :-) – 2017-01-12
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    @JefLaga, I have added a screenshot of the part of the page. – 2017-01-13

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If $G = \langle X |\, R_X \rangle$ and $H = \langle Y | \,R_Y\rangle$ are presentations of the groups $G$ and $H$ and if $\phi \colon H \rightarrow \text{Aut}(G) $ is a homomorphism, the corresponding semidirect product $G \rtimes_{\phi} H$ has the presentation $$\big\langle X\sqcup Y \,\mid\, R_X\sqcup R_Y\sqcup \{y^{-1}xy = \phi_y(x) \mid x \in X,\, y \in Y \} \big\rangle $$ Which is quite a pain to prove rigorously, but not too hard to grasp intuitively if one is familiar with semidirect products.

Apply this to $G = D_4$ and $H = \text{Aut}(D_4)$, while noting that this last group is isomorphic to $\text{Hol}(\mathbb{Z}/{4\mathbb{Z}})\cong \mathbb{Z}/{4\mathbb{Z}} \rtimes \mathbb{Z}/{2\mathbb{Z}}$ where this last group has the presentation $$\big\langle x,y \mid\,x^4 = y^2=1,\, yxy = x^3 \big\rangle$$ I hope this helps!