1. **State the problem:** We know that $y$ varies directly as $x$ and inversely as the square of $z$. Given $y=8$ when $x=4$ and $z=1$, find $y$ when $x=9$ and $z=3$. 2. **Write the variation formula:** Since $y$ varies directly as $x$ and inversely as $z^2$, the formula is: $$y = k \frac{x}{z^2}$$ where $k$ is the constant of proportionality. 3. **Find the constant $k$ using the given values:** Substitute $y=8$, $x=4$, and $z=1$: $$8 = k \frac{4}{1^2} = 4k$$ Solve for $k$: $$k = \frac{8}{4} = 2$$ 4. **Find $y$ for $x=9$ and $z=3$:** Use the formula with $k=2$: $$y = 2 \frac{9}{3^2} = 2 \frac{9}{9} = 2 \times 1 = 2$$ **Final answer:** $$y = 2$$