From this little example, I can only come to the conclusion to recommend to use another book. SCNR.
The idea of the dipole is to have a charge ##Q## at ##(0,0,d/2)=d/2 \vec{e}_3## and ##-Q## at ##(0,0,-d/2)=-d/2 \vec{e}_3##. The field is given by the superposition of the field of these point charges (Coulomb fields), i.e.,
$$\vec{E}=\frac{Q}{4 \pi \epsilon_0} \left [\frac{\vec{r}-d/2 \vec{e}_3}{|(\vec{r}-d/2 \vec{e}_3|^3}-\frac{\vec{r}+d/2 \vec{e}_3}{|\vec{r}+d/2 \vec{e}_3|^3} \right].$$
The next step usually is to make ##d \rightarrow 0## while keeping ##Q d=\text{const}##. This leads to the dipole approximation of the field for this charge configuration. This is a somewhat cumbersome calculation, however.
It's easier to use the electrostatic potential:
$$\Phi(\vec{r},d)=\frac{Q}{4 \pi \epsilon_0} \left [\frac{1}{|\vec{r}-d/2 \vec{e}_3|} - \frac{1}{|\vec{r}+d/2 \vec{e}_3|} \right].$$
Now ##|\vec{r} \pm d/2 \vec{e}_3|=\sqrt{x_1^2 +x_2^2 + (x_3\pm d/2)}^2##.
The Taylor expansion of ##\Phi## wrt. ##d## is
$$\Phi(\vec{r},d)=\Phi(\vec{r},0) + d \partial_d \Phi(\vec{r},0)+\mathcal{O}(d^2) = d \partial_d \Phi(\vec{r},0)+\mathcal{O}(d^2).$$
This gives
$$\Phi(\vec{r},d)=\frac{Q d x_3}{4 \pi \epsilon_0 r^3} + \mathcal{O}(Q d^2).$$
Defining the dipole moment
$$\vec{p}=Q d \vec{e}_3$$
Making now ##d \rightarrow 0## but keeping ##Q d=\text{const}## you get the dipole potential,
$$\Phi(\vec{r})=\frac{\vec{p} \cdot \vec{r}}{4 \pi \epsilon r^3}.$$
The field is
$$\vec{E}=-\vec{\nabla} \Phi=-\frac{\vec{p}}{4 \pi \epsilon_0 r^3} + \frac{3 (\vec{p} \cdot \vec{r}) \vec{r}}{4 \pi \epsilon_0 r^5} = \frac{1}{4 \pi \epsilon_0 r^5} [3 (\vec{r} \cdot \vec{p}) \vec{r}-r^2 \vec{p}].$$
