Answer
$$-1+\frac{8}{\pi}$$
Work Step by Step
Given
$$
f(x, y)=\ln x+\ln y, \quad \mathbf{r}(t)=\left\langle\cos t, t^{2}\right\rangle, \quad t=\frac{\pi}{4}
$$
since $\mathbf{r}(\pi/4)=\langle \frac{1}{\sqrt{2}},\frac{\pi^2}{16}\rangle$ and
$$
\begin{aligned}
\nabla f &=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle \\
&=\langle \frac{1}{x},\frac{1}{y}\rangle \\
\mathbf{r}^{\prime}(t) &=\left\langle -\sin t, 2t\right\rangle
\end{aligned}
$$
Then
\begin{aligned}
\left.\frac{d}{d t} f(\mathbf{r}(t))\right|_{t=\pi/4}&=\nabla f_{\mathbf{r}(\pi/4)} \cdot \mathbf{r}^{\prime}(\pi/4)\\
&=\langle\sqrt{2},\frac{16}{\pi^2}\rangle \cdot\langle -1/\sqrt{2},\pi/2\rangle \\
&=-1+\frac{8}{\pi}
\end{aligned}
