Using Fundamental Theorem of Algebra to find $z_0$ such that $|p(z_0)| < |p(0)|$ A question from the book Hubbard's Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach (5th edition):
I was able to easily find some point $z_0 = i/3$ (just by trying out points), but am unsure how to use their equation in their proof to find this point.

Consider the polynomial $p(z) = z^8 + z^4 + z^2 + 1$ where $z \in \mathbb{C}$, use the construction in the proof of the fundamental theorem of algebra, using equation 1.6.28 to find a point $z_0$ such that $|p(z_0)| < |p(0)| = 1$.


Here is a rough idea of how they prove The Fundamental Theorem of Algebra in their book:
First, they showed that for any monic polynomial of degree $k > 0$ with complex coefficients $p(z) = z^k + a_{k-1} z^{k-1} + \dots + a_0$, that $|p(z)|$ always has a global minimum at $z_0 \in \mathbb{C}$ for some $R>0$ with $|z_0| \leq R$.
Afterwards to show $p(z_0) = 0$, they assume instead $p(z_0) \neq 0$ to find a point $z$ such that $|p(z)| < |p(z_0)|$, which would lead to a  contradiction.
To do this, they let $z = z_0 + u$ so that:
$$\begin{align} p(z) &= (z_0 + u)^k + a_{k-1} (z_0 + u)^{k-1} + \dots + a_0 \\ &= u^k + b_{k-1}u^{k-1} + \dots + b_0 \\ &= q(u) \end{align}$$
where it can be shown $b_0 = z_0^k + a_{k-1}z_0^{k-1} + \dots + a_0 = p(z_0) \neq 0$
If we let $j>0$ be the smallest power of $q(u)$ with a non-zero coefficient, then we have equation 1.6.28:
$$\boxed{ q(u) = b_0 + b_j u^j + (b_{j+1} u^{j+1} + \dots + u^k) = p(z) = p(z_0 + u)}$$
Noting $u$ can be written as $\rho e^{i\theta}$, it can be imagined $b_0 + b_j u^j$ is travelling around a circle with center $b_0 = p(z_0)$. It can then be shown there exists a $\rho$ (I am not detailing its value here) such that $|b_j|\rho^j < |b_0|$,  so that for some values of $\theta$ we have $|b_0 + b_j u^j| < |b_0|$ (i.e. visually it will be a point on the line segment between $0$ and $b_0$)  and $|b_{j+1} u^{j+1} + \dots + u^k| < |b_j|\rho^j$ (the distance between the points $b_0 + b_ju^j$ and $b_0$). This leads to the contradiction $|p(z)| = |p(z_0 + u)| < |b_0| = |p(z_0)|$, since $|p(z_0)|$ is the minimum of the modulus of polynomial $p$.

I detailed the proof if it is needed (if you think I have left something out from the book let me know), but now I am uncertain how to use the equation 1.6.28 (boxed equation above) in their proof to find this point $z_0$. Any hints or ideas would be greatly appreciated.

In your case, you have $j=2$, $b_0=b_2=1$. Therefore $\rho=1/2$ satisfies the required inequality.
And for $\vert u \vert \lt 1/2$ you have
$$\vert u^4 +u^8 \vert \le \vert u \vert^4 + \vert u \vert^8 \le 2 \vert u \vert^4 \le 1/8 \lt \vert b_j \vert \rho =1/2$$
Applying the quoted theorem, take the point located on the circle of radius equal to $1/2$ and centered on the origin for which the modulus of $z^2 +1$ is minimum. That is $i/2$.
You’ll be able to verify that indeed $\vert p(i/2) \vert \lt 1$.
Note: take care that $z_0$ is used in the exercise and in the theorem for different things. That can be misleading.
