Answer
$\frac{dz}{dt} = 0$
Work Step by Step
\[
z = f(x,y), \qquad x = t^2,\qquad y = t^3
\]
We are given that \(f_x(4,8) = 3\) and \(f_y(4,8) = -1\).
To find \(\dfrac{dz}{dt}\) when \(t = 2\), we use the multivariable chain rule:
\[
\frac{dz}{dt} = f_x(x,y)\frac{dx}{dt} + f_y(x,y)\frac{dy}{dt}.
\]
Since
\[
x = t^2 \Rightarrow \frac{dx}{dt} = 2t, \qquad
y = t^3 \Rightarrow \frac{dy}{dt} = 3t^2,
\]
we substitute into the formula:
\[
\frac{dz}{dt} = f_x(x,y)(2t) + f_y(x,y)(3t^2).
\]
At \(t = 2\), we have \(x = 4\) and \(y = 8\), so
\[
\frac{dz}{dt}
= f_x(4,8)(2 \cdot 2) + f_y(4,8)(3 \cdot 2^2)
= 3(4) + (-1)(12)
= 12 - 12
= 0.
\]
\[
\boxed{\frac{dz}{dt} = 0}
\]
