I know what the chain rule is but i am unsure how to apply it to this case.
$$z=x^3\cdot e^{2y}$$
$$x=2t$$
$$y=t^2$$
Edit: Fixed title to $\frac{dz}{dt}$.
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$\begingroup$Note the multivariable chain rule: $$\frac{dz}{dt}=\frac{\partial z}{\partial x}\cdot \frac{dx}{dt}+\frac{\partial z}{\partial y}\cdot \frac{dy}{dt} \tag{1}$$ Simply evaluate the ordinary and partial derivatives, and substitute them into the equation above. You should have a result for $\frac{dz}{dt}$ in terms of $x,y$ and $t$. Then, substitute the parametrizations for $x(t)$ and $y(t)$ to obtain an expression for $\frac{dz}{dt}$ strictly in terms of $t$ as done on the first example of the link I provided.
Let's start by evaluating $\frac{\partial z}{\partial x}$. Since we must treat the variable $y$ as a constant as a result of partial differentiation, $e^{2y}$ is also treated as one. Therefore: $$\frac{\partial z}{\partial x}=3x^2\cdot e^{2y}$$ Can you continue?
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