Answer
$$\frac{\partial z}{\partial s} = xy^2 (2ycos \space t + 3xsin \space t) $$ $$\frac{\partial z}{\partial t} = sxy^2 (-2y \space sin \space t + 3x \space cos \space t) $$
Work Step by Step
According to the Chain Rule:
$$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}$$
$$\frac{\partial z}{\partial x} = (x^2y^3)' = 2xy^3$$ $$\frac{\partial x}{\partial s} = (s \space cos(t))' = cos(t)$$ $$\frac{\partial z}{\partial y} = (x^2y^3)' = 3x^2y^2$$ $$\frac{\partial y}{\partial s} = (s \space sin(t))' = sin(t)$$
$$\frac{\partial z}{\partial s} = (2xy^3)(cos \space t) + (3x^2y^2)(sin \space t)$$ $$= xy^2 (2ycos \space t + 3xsin \space t) $$
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$$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}$$
$$\frac{\partial x}{\partial t} = (s \space cos(t))' = -s \space sin(t)$$ $$\frac{\partial y}{\partial t} = (s \space sin(t))' = s \space cos(t)$$
$$\frac{\partial z}{\partial t} = (2xy^3)(-s \space sin \space t) + (3x^2y^2)(s \space cos \space t)$$ $$= sxy^2 (-2y \space sin \space t + 3x \space cos \space t) $$
