Answer
$$-56$$
Work Step by Step
Given
$$
f(x, y)=x-x y, \quad \mathbf{r}(t)=\left\langle t^{2}, t^{2}-4 t\right\rangle, \quad t=4
$$
since $\mathbf{r}(4)=\langle 16,0\rangle$ and
$$
\begin{aligned}
\nabla f &=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle \\
&=\langle1-y,-x\rangle \\
\mathbf{r}^{\prime}(t) &=\left\langle 2t, 2t-4\right\rangle
\end{aligned}
$$
Then
\begin{aligned}
\left.\frac{d}{d t} f(\mathbf{r}(t))\right|_{t=4}&=\nabla f_{\mathbf{r}(4)} \cdot \mathbf{r}^{\prime}(4)\\
&=\langle1,-16\rangle \cdot\langle 8,4\rangle \\
&= 8-64=-56
\end{aligned}
