Answer
$$-72$$
Work Step by Step
Given $$ f(x, y)=3 x-7 y, \quad \mathbf{r}(t)=\left\langle t^{2}, t^{3}\right\rangle, \quad t=2$$
Since $ \mathbf{r}(2)=\langle4,8\rangle$ and
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\
&=\langle 3,-7\rangle\\
\mathbf{r}^{\prime}(t)&=\langle2 t, 3 t^2\rangle
\end{align*}
Then
\begin{align*}
\frac{d}{d t} f(\mathbf{r}(t))&=\nabla f_{\mathbf{r}(t)} \cdot \mathbf{r}^{\prime}(t)\\
&= \langle 3,-7\rangle\cdot \langle2 t, 3 t^2\rangle
\end{align*}
Hence
\begin{align*}
\frac{d}{d t} f(\mathbf{r}(t))\bigg|_{t=2} &= \langle 3,-7\rangle\cdot \langle4, 12\rangle \\
&=12-84=-72
\end{align*}
