Answer
$$\frac{dz}{dt}=\frac{6\pi}{(x+2y)^{2}}=\frac{6\pi}{(e^{\pi t}+2e^{-\pi t})^{2}}$$
Work Step by Step
By the Chain rule, we have
$$\frac{dz}{dt}=\frac{\partial{z}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{z}}{\partial{y}}\frac{dy}{dt}$$
$$\Rightarrow\frac{dz}{dt}=\frac{(1)(x+2y)-(x-y)(1)}{(x+2y)^{2}}\cdot(\pi e^{\pi t})+\frac{(-1)(x+2y)-(x-y)(2)}{(x+2y)^{2}}\cdot(-\pi e^{-\pi t})$$
$$\Rightarrow\frac{dz}{dt}=\frac{3y\cdot(\pi e^{\pi t})}{(x+2y)^{2}}+\frac{(-3x)\cdot(-\pi e^{-\pi t})}{(x+2y)^{2}}=\frac{3\pi}{(x+2y)^{2}}(ye^{\pi t}+xe^{-\pi t})$$
Plug in $x=e^{\pi t}$ and $y=e^{-\pi t}$, therefore we obtain
$$\frac{dz}{dt}=\frac{3\pi}{(e^{\pi t}+2e^{-\pi t})^{2}}(e^{-\pi t}\cdot e^{\pi t}+e^{\pi t}\cdot e^{-\pi t})=\frac{3\pi}{(e^{\pi t}+2e^{-\pi t})^{2}}(1+1)=\frac{6\pi}{(e^{\pi t}+2e^{-\pi t})^{2}}$$
