If $f(x,y)=4-x^{2}-2y^{2}$, find $f_{x}(1,1)$ and $f_{y}(1,1)$ and interpret these numbers as slopes.
I understand that this is a partial derivative
and I can solve it by setting $x$ or $y$ as constant.
by setting $y$ as constant, then how come the vector for slope is
$\mathbf{r}(t)=\langle 1+t,1,1-2t \rangle$ for $T_{1}$ in this case as indicated in the picture?
?
Back up a bit. The first $\mathbf{r}(t)$ in that paragraph is a vector equation for $C_1$, that is, the curve defined by equations $y=1$ and $z=f(x,y)$. We could translate it into parametric equations as $x=t$, $y=1$, and $z=f(t,1) = 2-t^2$.
$T_1$ is the line tangent to $C_1$ at the point $P=(1,1,2)$. With the vector equation $\mathbf{r}(t)$ used above, $T_1$ has direction vector $\mathbf{v}$ given by $\mathbf{r}'(1)$. Since $\mathbf{r}'(t) = \left<1,0,-2t\right>$, we have $\mathbf{v} = \left<1,0,-2\right>$.
The second $\mathbf{r}(t)$ is a vector equation for $T_1$. We can construct this using the point $P$ and the direction vector $\mathbf{v}$. Setting $\mathbf{r}_0 = \overrightarrow{OP} = \left<1,1,2\right>$, $T_1$ has vector equation $$ \mathbf{r}(t) = \mathbf{r}_0 + t \mathbf{v} = \left<1,1,2\right>+t\left<1,0,-2\right> = \left<1+t,1,1-2t\right> $$
It's perhaps unfortunate that Stewart uses $\mathbf{r}(t)$ as a generic variable for vector functions, rather than a specific named one. But it's akin to giving parametric equations for $C_1$ as $x=t$, $y=1$, and $z=2-t^2$, then immediately after giving parametric equations for $T_1$ as $x=1+t$, $y=1$, and $z=1-2t$. The $x$, $y$, and $z$ in each set of equations are just variable names.
Treating $y$ as a constant, you'll find the partial derivative $f_x(x,y)=-2x$, which at the given point evaluates to $f_x(1,1)=-2$. Also, note that the given point has coordinates $x=1$, $y=1$, and $z=4-1^2-2\cdot1^2=1$, i.e. $(1,1,1)$.
Geometrically, setting $y=1$ represents a plane parallel to the $xz$-plane. When this plane intersects the given surface, it creates a certain curve as the cross-section — that's what $C_1$ is. The partial derivative $f_x(x,y)$ can be interpreted as the slope of the tangent line to this curve inside this plane. (I bet all of this was illustrated in the textbook with pictures before the example and in this example.)
Informally speaking, you only have $x,z$ coordinates in this vertical plane, since $y$ has been fixed to be $y=1$. Inside this plane:
The equation of the curve $C_1$ that is the cross-section is $z=4-x^2-2\cdot1^2=2-x^2$ (back to technically speaking, along with $y=1$).
The equation of tangent line $T_1$ to the curve is $z-1=-2(x-1)$ or $z=-2x+3$, again along with $y=1$. We can parametrize it by letting $x-1=t$, so $x=1+t$ and $z=1-2t$. And don't forget that $y=1$.
So the entire parametrization is $x=1+t$, $y=1$, $z=1-2t$, or in vector form $\mathbf{r}(t)=\langle 1+t,1,1-2t \rangle$.