Answer
$$\frac{\partial y}{\partial s} = e^{x+2y} (1/t - 2t/s^2) $$
$$\frac{\partial z}{\partial s} = e^{x+2y} (-s/t^2 + 2/s) $$
Work Step by Step
According to the Chain Rule:
$$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}$$
$$\frac{\partial z}{\partial x} = (e^{x+2y})' = e^{x+2y}$$ $$\frac{\partial x}{\partial s} = (s/t)' = 1/t$$ $$\frac{\partial z}{\partial y} = (e^{x+2y})' = 2e^{x+2y}$$ $$\frac{\partial y}{\partial s} = (t/s)' = -t/s^2$$
$$\frac{\partial z}{\partial s} = (e^{x+2y})(1/t) + (2e^{x+2y})(-t/s^2)$$ $$= e^{x+2y} (1/t - 2t/s^2) $$
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$$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}$$
$$\frac{\partial x}{\partial t} = (s/t)' = -s/t^2$$ $$\frac{\partial y}{\partial t} = (t/s)' = 1/s$$
$$\frac{\partial z}{\partial s} = (e^{x+2y})(-s/t^2) + (2e^{x+2y})(1/s)$$ $$= e^{x+2y} (-s/t^2 + 2/s) $$
