1. **State the problem:** We are given the differential equation $$y'(t) = \frac{1}{t} y(t) - y(t)^2$$ and the substitution $$y(t) = \frac{1}{z(t)}.$$ We want to find an equation for $z(t)$. 2. **Use the substitution:** Since $$y(t) = \frac{1}{z(t)},$$ then by the chain rule, $$y'(t) = -\frac{z'(t)}{z(t)^2}.$$ 3. **Rewrite the original equation using $z(t)$:** Substitute $y(t)$ and $y'(t)$ into the original equation: $$-\frac{z'(t)}{z(t)^2} = \frac{1}{t} \cdot \frac{1}{z(t)} - \left(\frac{1}{z(t)}\right)^2 = \frac{1}{t z(t)} - \frac{1}{z(t)^2}.$$ 4. **Multiply both sides by $z(t)^2$ to clear denominators:** $$-z'(t) = \frac{z(t)^2}{t z(t)} - \frac{z(t)^2}{z(t)^2} = \frac{z(t)}{t} - 1.$$ 5. **Rewrite the equation for $z'(t)$:** $$z'(t) = 1 - \frac{z(t)}{t}.$$ 6. **Final result:** The differential equation for $z(t)$ is $$\boxed{z'(t) = 1 - \frac{z(t)}{t}}.$$