To find $\frac{dz}{dt}$ using the chain rule, let's go through each step carefully.

Given: $z = (x + y)e^y$ where $x = 3t$ and $y = 1 - t^2$.

Step 1: Differentiate $z$ with respect to $t$ using the chain rule

Since $z$ is a function of both $x$ and $y$, which are in turn functions of $t$, we apply the chain rule as follows: $\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}$

Step 2: Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$

1. Compute $\frac{\partial z}{\partial x}$: $\frac{\partial z}{\partial x} = e^y$

2. Compute $\frac{\partial z}{\partial y}$: [ \frac{\partial z}{\partial y} = (x + y)e^y \cdot \frac{d}{dy}(y) + e^y \cdot \frac{d}{dy}(x + y)