If [tex]x[/tex] varies directly as [tex]y^2[/tex], then there is some constant [tex]a[/tex] for which
[tex]x=ay^2[/tex]
Similarly, there is some constant [tex]b[/tex] such that
[tex]y=bz^3[/tex]
Given that [tex]x=-16[/tex] when [tex]z=2[/tex], we have
[tex]\begin{cases}-16=ay^2\\y=8b\end{cases}\implies-16=a(8b)^2\implies ab^2=-\dfrac14[/tex]
Now when [tex]z=\frac12[/tex], we get
[tex]\begin{cases}x=ay^2\\y=\frac b8\end{cases}\implies x=a\left(\dfrac b8\right)^2=\dfrac{ab^2}{64}=\dfrac{-\frac14}{64}=\boxed{-\dfrac1{256}}[/tex]
