Answer
$$\frac{dz}{dt} = \frac{-ye^t + xe^{-t}}{x^2 + y^2}$$
Work Step by Step
According to the Chain Rule:
$$\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} +\frac{\partial z}{\partial y} \frac{dy}{dt} $$
$$ \frac{\partial z}{\partial x} = \frac{\partial (tan^{-1}(y/x))}{\partial x} = \Bigg(\frac{1}{1 + (y/x)^2}\Bigg) (-yx^{-2})$$ $$\frac{dx}{dt} = \frac{d(e^t)}{dt} = e^t$$ $$\frac{\partial z}{\partial y} = \frac{\partial (tan^{-1}(y/x))}{\partial y} =\Bigg(\frac{1}{1 + (y/x)^2}\Bigg) (x^{-1})$$ $$ \frac{dy}{dt} = \frac{d(1 - e^{-t})}{dt} = (-e^{-t})(-1) = e^{-t}$$
Therefore:
$$\frac{dz}{dt} =\Bigg(\frac{1}{1 + (y/x)^2}\Bigg) ( (-yx^{-2}) (e^t)+ \Bigg(\frac{1}{1 + (y/x)^2}\Bigg) (x^{-1}) (e^{-t})$$ $$\frac{dz}{dt} =\Bigg(\frac{1}{1 + (y/x)^2}\Bigg) \Bigg (-yx^{-2}e^t + x^{-1}e^{-t} \Bigg)$$
** Multiply the fraction by $\frac{x^2}{x^2}$
$$\frac{dz}{dt} =\Bigg(\frac{x^2}{x^2 + y^2}\Bigg) \Bigg (-yx^{-2}e^t + x^{-1}e^{-t} \Bigg)$$ $$\frac{dz}{dt} = \frac{-ye^t + xe^{-t}}{x^2 + y^2}$$
