Answer
$$-3$$
Work Step by Step
Given $$ f(x, y)=x^{2}-3 x y, \quad \mathbf{r}(t)=\langle\cos t, \sin t\rangle, \quad t=0$$
Since $ \mathbf{r}(0)=\langle 1,0 \rangle$ and
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\
&=\langle 2x-3y, -3x\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 2\cos t-3\sin t, -3\cos t\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 2,-3\rangle\cdot \langle0, 1\rangle \\
&= -3
\end{align*}
