Answer
$$-7$$
Work Step by Step
Given $$ f(x, y)=3 x-7 y, \quad \mathbf{r}(t)=\langle\cos t, \sin t\rangle, \quad t=0$$
Since $ \mathbf{r}(0)=\langle1,0\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)&=\langle-\sin t, \cos t\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 \langle-\sin t, \cos t\rangle
\end{align*}
Hence
\begin{align*}
\frac{d}{d t} f(\mathbf{r}(t))\bigg|_{t=0} &= \langle 3,-7\rangle\cdot \langle0, 1\rangle \\
&=-7
\end{align*}
