1. **State the problem:** We need to find the derivative of the function given implicitly as $$\frac{dt}{dy} = k y_0^{e^{kt}}$$ with respect to $y$. 2. **Rewrite the problem:** The expression $$\frac{dt}{dy}$$ means the derivative of $t$ with respect to $y$. We want to find $$\frac{d}{dy}\left(\frac{dt}{dy}\right)$$ or equivalently $$\frac{d^2 t}{dy^2}$$. 3. **Identify the function:** The right side is $$k y_0^{e^{kt}}$$, where $k$ and $y_0$ are constants, and $t$ is implicitly a function of $y$. 4. **Apply the chain rule:** To differentiate $$\frac{dt}{dy} = k y_0^{e^{kt}}$$ with respect to $y$, we write $$\frac{d}{dy}\left(\frac{dt}{dy}\right) = \frac{d}{dy}\left(k y_0^{e^{kt}}\right) = k \frac{d}{dy}\left(y_0^{e^{kt}}\right)$$ 5. **Differentiate the exponential expression:** Recall that $$y_0^{e^{kt}} = e^{e^{kt} \ln y_0}$$. Using the chain rule, $$\frac{d}{dy} y_0^{e^{kt}} = y_0^{e^{kt}} \cdot \frac{d}{dy} (e^{kt} \ln y_0) = y_0^{e^{kt}} \ln y_0 \cdot \frac{d}{dy} e^{kt}$$ 6. **Differentiate $$e^{kt}$$ with respect to $y$:** Since $t$ depends on $y$, by chain rule, $$\frac{d}{dy} e^{kt} = e^{kt} \cdot k \frac{dt}{dy}$$ 7. **Substitute back:** $$\frac{d}{dy} y_0^{e^{kt}} = y_0^{e^{kt}} \ln y_0 \cdot e^{kt} k \frac{dt}{dy}$$ 8. **Put it all together:** $$\frac{d^2 t}{dy^2} = k \cdot y_0^{e^{kt}} \ln y_0 \cdot e^{kt} k \frac{dt}{dy} = k^2 y_0^{e^{kt}} e^{kt} \ln y_0 \frac{dt}{dy}$$ 9. **Recall $$\frac{dt}{dy} = k y_0^{e^{kt}}$$, substitute it back: $$\frac{d^2 t}{dy^2} = k^2 y_0^{e^{kt}} e^{kt} \ln y_0 \cdot k y_0^{e^{kt}} = k^3 e^{kt} \ln y_0 \left(y_0^{e^{kt}}\right)^2$$ **Final answer:** $$\boxed{\frac{d^2 t}{dy^2} = k^3 e^{kt} \ln y_0 \left(y_0^{e^{kt}}\right)^2}$$