#### Answer

$$15\sin(6)$$

#### Work Step by Step

Given $$ f(x, y)=\cos (y-x), \quad \mathbf{r}(t)=\left\langle e^{t}, e^{2 t}\right\rangle, \quad t=\ln 3$$
Since $ \mathbf{r}(\ln 3)=\langle 3,9 \rangle$ and
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\
&=\langle \sin (y-x),- \sin(y-x) \rangle\\
\mathbf{r}^{\prime}(t)&=\langle e^t,2e^{2 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 \sin (-6), -\sin (-6)\rangle\cdot\langle e^t,2e^{2 t} \rangle
\end{align*}
Hence
\begin{align*}
\frac{d}{d t} f(\mathbf{r}(t))\bigg|_{t=\ln 3} &= \langle - \sin (6), \sin (6)\rangle\cdot \langle3,18\rangle \\
&=15\sin(6)
\end{align*}