Find $y \in K$, with $K = F(y)$, such that $y^3 \in F$. Let $K / F$ be a monogenic extension of fields such that the characteristic of $F$ is not $3$ and $[K : F] = 3$. Can I find $y \in K$, with $K = F(y)$, such that $y^3 \in F$? If we replace $3$ by $2$ in the previous paragraph, the answer is yes: as $K / F$ is monogenic, there exists $x \in K$ such that $K = F(x)$. Thus, $\\{1 , x\\}$ is a basis for the $F$-vector space $K$ and $x^2 \in F(x)$. Then there exist $\alpha , \beta \in F$ such that $x^2 = \alpha + \beta x$. Now, the element $y = \frac{\beta}{2} - x$ makes sense (as the characteristic of $F$ is not $2$) and satisfies that $K = F(y)$ and $y^2 \in F$.

No, you cannot do it for extensions like $\Bbb Q(\cos(2\pi/9))\Bbb Q$. All finite extensions in characteristic zero are monogenic. This extension is totally real, and of degree three. A monogenic (pure) cubic extension of $\Bbb Q$ has the form $\Bbb Q(\sqrt[3]a)$ and is not totally real. The element $\sqrt[3]a$ has a conjugate $\exp(2\pi i/3)\sqrt[3]a$ which is not real.