#### Answer

$$5\cos (1)$$

#### Work Step by Step

Given $$ f(x, y)=\sin (x y), \quad \mathbf{r}(t)=\left\langle e^{2 t}, e^{3 t}\right\rangle, \quad t=0$$
Since $ \mathbf{r}(0)=\langle 1,1 \rangle$ and
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\
&=\langle y\cos (xy), x\cos (xy)\rangle\\
\mathbf{r}^{\prime}(t)&=\langle 2e^{2 t}, 3e^{3 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 \cos (1), \cos (1)\rangle\cdot\langle 2e^{2 t}, 3e^{3 t} \rangle
\end{align*}
Hence
\begin{align*}
\frac{d}{d t} f(\mathbf{r}(t))\bigg|_{t=\pi/2} &= \langle \cos (1), \cos (1)\rangle\cdot \langle2,3\rangle \\
&= 5\cos (1)
\end{align*}