Let $f(x, y)$ denote a function of two variables whose partial derivatives – $f_x(x, y)$ and $f_y(x, y)$ – are considered to be “known” functions. Let $A(x, y)$ be another function whose partial derivatives, $A_x(x, y)$ and $A_y(x, y)$, are also considered to be known. Now consider the function $F(t) = f(A(4t, t), A(t, 2t))$. Compute, in terms of the known quantities, evaluated at appropriate arguments, the derivative $F'(t)$.
Progress
I know this question is very ambiguous, the professor seems give us questions like these a lot. I'm assuming you take the function $F(t) = f(A(4t, t), A(t, 2t))$ and you have to use the chain rule on it, but I am confused as into how to take the derivative of the function and then answer it in terms the "known" variables.
You will have to use the multivariate chain rule, like you said, but you will have to apply the rule multiple times to get the desired result:
The multivariate chain rule goes like this:
$\frac{df}{dt} = \frac{\partial f}{\partial x_1}\times\frac{d x_1}{dt}+\cdots+\frac{\partial f}{\partial x_n}\times\frac{d x_n}{dt}$
For your functions:
\begin{align} F'(t)&=f_x(A(4t,t),A(t,2t))\times\frac{dA(4t,t)}{dt}+f_y(A(4t,t),A(t,2t))\times\frac{dA(t,2t)}{dt}\\&=f_x(A(4t,t),A(t,2t))\times(A_x(4t,t)\times4+A_y(4t,t))\\&+f_y(A(4t,t),A(t,2t))\times(A_x(t,2t)+A_y(t,2t)\times2) \end{align}
When you are evaluating $\frac{dA(4t,t)}{dt}$ and $\frac{dA(t,2t)}{dt}$ that you got from the first chain rule, you have to apply the rule again.