Answer
$$\frac{dz}{dt}=\frac{6\pi}{(e^{\pi t}+2e^{-\pi t})^2}$$
Work Step by Step
$$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}=\frac{\partial}{\partial x}\Big(\frac{x-y}{x+2y}\Big)\frac{d}{dt}(e^{\pi t})+\frac{\partial }{\partial y}\Big(\frac{x-y}{x+2y}\Big)\frac{d}{dt}(e^{-\pi t})=
\frac{\frac{\partial}{\partial x}(x-y)\cdot(x+2y)-(x-y)\frac{\partial}{\partial x}(x+2y)}{(x+2y)^2}\cdot\pi e^{\pi t}+
\frac{\frac{\partial}{\partial y}(x-y)\cdot(x+2y)-(x-y)\frac{\partial}{\partial y}(x+2y)}{(x+2y)^2}\cdot(-\pi e^{-\pi t})=
\frac{1\cdot(x+2y)-(x-y)\cdot1}{(x+2y)^2}\cdot\pi e^{\pi t}-\frac{-1\cdot(x+2y)-(x-y)\cdot2}{(x+2y)^2}\cdot\pi e^{-\pi t}=
\frac{3y}{(x+2y)^2}\cdot\pi e^{\pi t}-\frac{-3x}{(x+2y)^2}\cdot\pi e^{-\pi t}=
3\pi\frac{ye^{\pi t}+xe^{-\pi t}}{(x+2y)^2}$$
Expressing this in terms of $t$ we have:
$$\frac{dz}{dt}=3\pi\frac{ye^{\pi t}+xe^{-\pi t}}{(x+2y)^2}=
3\pi\frac{e^{-\pi t}\cdot e^{\pi t}+e^{\pi t}\cdot e^{-\pi t}}{(e^{\pi t}+2e^{-\pi t})^2}=\frac{6\pi}{(e^{\pi t}+2e^{-\pi t})^2}$$
