So this is a chain rule question. So I drew out that I need $\frac{dz}{dt}=\frac{dz}{dx}\frac{dx}{dt}+\frac{dz}{dy}\frac{dy}{dt}$. I'm not sure if this is correct, but from here I computed that $\frac{dz}{dx}= x$, $\frac{dx}{dt}=7$,$\frac{dz}{dy}=ye^y$, and $\frac{dy}{dt}=(-2t)$. I then got $\frac{dz}{dt}=7x+ye^y(-2t)$, but this is not correct.
I still get confused on these and partial derivatives, on which variable is the constant and how to compute these.
On
Your mistake lies in how you computed the derivatives of $z$ with respect to either $x$ or $y$. You should have
$$z=(x+y)e^y\implies\begin{cases}\frac{\partial z}{\partial x}=e^y\\\frac{\partial z}{\partial y}=e^y+(x+y)e^y=(x+y+1)e^y\end{cases}$$
Then when you simplify everything to find $\frac{\mathrm dz}{\mathrm dt}$, remember to replace every instance of $x$ and $y$ with their corresponding expressions in terms of $t$.
$$\frac{dz}{dx}= e^y$$ $$\frac{dz}{dy}= xe^y+e^y+ye^y$$