y varies directly as x and inversely as the square of z means:
[tex]y=\frac{kx}{z^2}[/tex]where k is some constant. Substituting with y = 16, x = 32, and z = 4, and solving for k:
[tex]\begin{gathered} 16=\frac{k\cdot32}{4^2} \\ \frac{16\cdot4^2}{32}=k \\ 8=k \end{gathered}[/tex]Substituting with k = 8, x = 3, and z = 6, and solving for y:
[tex]\begin{gathered} y=\frac{8\cdot3}{6^2} \\ y=\frac{2}{3} \end{gathered}[/tex]