Answer
At point $A$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right)$ is zero.
At point $B$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right)$ is negative.
At point $C$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right)$ is positive.
At point $D$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right)$ is zero.
Work Step by Step
By definition the derivative $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right)$ is given by
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta f}}{{\Delta t}}$
Since we traverse in the increasing of $t$, $\Delta t > 0$. So, the sign depends on the sign of $\Delta f$.
Point $A$:
At $A$, we see from Figure 15, the path of the particle touches the level curve $0$ for some time, so $\Delta f = 0$. Thus, at point $A$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right)$ is zero.
Point $B$:
At $B$, we see from Figure 15, the particle moves from the level curve $-10$ to the level curve $-20$, so $\Delta f < 0$. Thus, at point $B$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right)$ is negative.
Point $C$:
At $C$, we see from Figure 15, the particle moves from the level curve $-10$ to the level curve $0$, so $\Delta f > 0$. Thus, at point $C$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right)$ is positive.
Point $D$:
At $D$, we see from Figure 15, the path of the particle touches the level curve $30$ for some time, so $\Delta f = 0$. Thus, at point $D$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right)$ is zero.
