So I started off with $z = x+iy$ where $x$ and $y$ are reals.
$\left|\operatorname{Re}\left((3+i+2(x-iy)(x-iy)-i(x+iy)\right)\right|$
$\left|\operatorname{Re}(3+i+2(x^2-2iyx-y^2)-ix+y)\right|$
$\left|\operatorname{Re}(3+i+2x^2-4iyx-2y^2-ix+y)\right|$
$\left|3+2x^2-2y^2+y\right|$
$3+2x^2-2y^2+y$
And then I'm not really sure where to go. Pretty sure I took the wrong approach because not using the initial information that $|z| \le 1$
On
You need to prove $\left|3+2x^2-2y^2+y\right|\le 6$ subject to $|z|\le 1 \iff x^2+y^2\le 1$.
Note: $$\left|3+2x^2-2y^2+y\right|=\left|3+2(x^2+y^2)-4y^2+y\right|\le |5-4y^2+y|=|4y^2-y-5|\le6 \Rightarrow \\ \begin{cases}4y^2-y-5\ge -6\\ 4y^2-y-5\le 6\end{cases} \Rightarrow \begin{cases}4y^2-y+1\ge 0\\ 4y^2-y-11\le0\end{cases} \Rightarrow \begin{cases}y\in(-\infty,+\infty)\\ y\in\left[\frac{1-\sqrt{177}}{8}\approx -1.538,\frac{1+\sqrt{177}}{8}\approx 1.788\right]\end{cases}.$$ Also note: $$x^2+y^2\le 1 \Rightarrow -1\le y\le 1.$$
On
Note that $3+2x^2-2y^2+y\leq5+y-4y^2$ if $x^2+y^2\leq 1$. Therefore, $$3+2x^2-2y^2+y\leq 5+y-4y^2=5+\frac{4y(1-4y)}{4}\leq 5+\frac{1}{4}\,\left(\frac{4y+(1-4y)}{2}\right)^2=\frac{81}{16}$$ for $y\in[-1,+1]$, where we have applied the AM-GM Inequality on the right-hand inequality. We also have $3+2x^2-2y^2+y\geq 3+y-2y^2$, so $$3+2x^2-2y^2+y\geq 3+y-2y^2=(3-2y)(1+y)\geq 0\text{ for all }y\in[-1,+1]$$ This shows that $$0\leq \text{Re}\big(3+\text{i}+2\,\bar{z}^2-\text{i}\,z\big)\leq \frac{81}{16}$$ for all complex numbers $z$ such that $|z|\leq 1$. The left-hand side is an equality iff $z=-\text{i}$, and the right-hand side is an equality iff $z=\dfrac{\pm3\sqrt{7}+\text{i}}{8}$
Note that $\operatorname{Re}(3+i+2\overline z^2-iz)=\operatorname{Re}(3+2\overline z^2-iz)$ and that therefore\begin{align}\bigl\lvert\operatorname{Re}(3+i+2\overline z^2-iz)\bigr\rvert&\leqslant\bigl\lvert3+2\overline z^2-iz\bigr\rvert\\&\leqslant3+2|z|^2+|z|\\&\leqslant6,\end{align}since $|z|\leqslant1$.