1. **State the problem:** Given $$z = e^x \sin y$$ with $$x = s t^2$$ and $$y = s^2 t$$, find the partial derivatives $$\frac{\partial z}{\partial s}$$ and $$\frac{\partial z}{\partial t}$$ using the chain rule. 2. **Recall the chain rule for multivariable functions:** $$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}$$ $$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}$$ 3. **Compute the partial derivatives of $$z$$ with respect to $$x$$ and $$y$$:** $$\frac{\partial z}{\partial x} = e^x \sin y$$ (since derivative of $$e^x$$ is $$e^x$$ and $$\sin y$$ is treated as constant) $$\frac{\partial z}{\partial y} = e^x \cos y$$ (derivative of $$\sin y$$ is $$\cos y$$) 4. **Compute the partial derivatives of $$x$$ and $$y$$ with respect to $$s$$ and $$t$$:** $$\frac{\partial x}{\partial s} = t^2$$ $$\frac{\partial x}{\partial t} = 2 s t$$ $$\frac{\partial y}{\partial s} = 2 s t$$ $$\frac{\partial y}{\partial t} = s^2$$ 5. **Substitute all into the chain rule formulas:** $$\frac{\partial z}{\partial s} = e^x \sin y \cdot t^2 + e^x \cos y \cdot 2 s t = e^{s t^2} \sin(s^2 t) t^2 + 2 s t e^{s t^2} \cos(s^2 t)$$ $$\frac{\partial z}{\partial t} = e^x \sin y \cdot 2 s t + e^x \cos y \cdot s^2 = 2 s t e^{s t^2} \sin(s^2 t) + s^2 e^{s t^2} \cos(s^2 t)$$ 6. **Final answers:** $$\boxed{\frac{\partial z}{\partial s} = e^{s t^2} \sin(s^2 t) t^2 + 2 s t e^{s t^2} \cos(s^2 t)}$$ $$\boxed{\frac{\partial z}{\partial t} = 2 s t e^{s t^2} \sin(s^2 t) + s^2 e^{s t^2} \cos(s^2 t)}$$