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