1. **State the problem:** We are given that $y$ varies directly as $x$ and inversely as the square of $z$. This means the relationship can be written as: $$y = \frac{kx}{z^2}$$ where $k$ is the constant of variation. 2. **Given values:** $y=8$, $x=4$, and $z=1$. 3. **Substitute the known values into the formula:** $$8 = \frac{k \times 4}{1^2} = 4k$$ 4. **Solve for $k$:** $$k = \frac{8}{4} = 2$$ 5. **Conclusion:** The constant of variation $k$ is 2.