Given that,
- $x^{2018} y^{2017} = \frac {1}{2} \quad \longrightarrow (1)$
- $x^{2016} y^{2019} =8 \quad \longrightarrow (2)$
Divide the equation $(1)$ by equation $(2),$ we get
$ \frac {x^{2018} y^{2017} }{x^{2016} y^{2019}} = \frac{ \frac{1}{2}}{8}$
$ \Rightarrow \frac {x^{2}}{y^{2}} = \frac {1}{16}$
$ \Rightarrow \boxed {\frac {x}{y} = \pm \frac{1}{4}}$
$ \Rightarrow \boxed {x = \pm \frac{y}{4}} \quad \longrightarrow (3)$
Put the value of $x$ in equation $(2),$ we get
$x^{2016} y^{2019} =8$
$ \Rightarrow \left( \pm \frac{y}{4} \right)^{2016} y^{2019} = 8$
$ \Rightarrow y^{2016} \ast y^{2019} = 4^{2016} \ast 8$
$ \Rightarrow y^{4035} = \left( 2^{2} \right)^{2016} \ast 8$
$ \Rightarrow y^{4035} = 2^{4032} \ast 2^{3}$
$ \Rightarrow y^{4035} = 2^{4035}$
$ \Rightarrow \boxed {y=2}$
Put the value of $y$ in equation $(3),$ we get
$ x = \pm \frac {y}{4} = \pm \frac {2}{4}$
$ \boxed {x= \pm \frac {1}{2}}$
Now, $ x^{2} + y^{3} = \left( \pm \frac {1}{2} \right)^{2} + (2)^{3}$
$ = \dfrac {1}{4} + 8$
$ = \dfrac {1+32}{4}$
$ = \dfrac {33}{4} $
Correct Answer $: \text{D}$