Answer
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right){|_{t = - 2}} = 197$
Work Step by Step
We have $f\left( {x,y,z} \right) = xz - {y^2}$ and ${\bf{r}}\left( t \right) = \left( {t,{t^3},1 - t} \right)$.
So,
$\nabla f = \left( {z, - 2y,x} \right)$, ${\ \ \ \ }$ $\nabla {f_{{\bf{r}}\left( t \right)}} = \left( {1 - t, - 2{t^3},t} \right)$
${\bf{r}}'\left( t \right) = \left( {1,3{t^2}, - 1} \right)$
At $t=-2$, we get $\nabla {f_{{\bf{r}}\left( { - 2} \right)}} = \left( {3,16, - 2} \right)$ and ${\bf{r}}'\left( { - 2} \right) = \left( {1,12, - 1} \right)$.
Using the Chain Rule for Paths (Theorem 2 of Section 15.5):
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right){|_{t = - 2}} = \nabla {f_{{\bf{r}}\left( { - 2} \right)}}\cdot{\bf{r}}'\left( { - 2} \right)$
$ = \left( {3,16, - 2} \right)\cdot\left( {1,12, - 1} \right)$
$ = 197$
