Determining Sign of Directional Derivate Graphically CONTEXT: First year uni calculus course question

How would you determine, for $f(x,y)$, $g(x,y)$ and $h(x,y)$, whether the directional derivatives at the indicated points are $>0$, $<0$ or $=0$ in the directions of $v=(1,2)$ and $w=(2,1)$?
For example, I can tell $D_{v}f(x,y)=0$ since it is tangent to the curve (and perpendicular to $\nabla f$ making the dot product $0$). But I am unsure about the five other directional derivatives.
Any help would be greatly appreciated.
 A: Given $z=f(x, y)$,
The directional derivative of $f$ at a some arbitrary point $P_0(x_0,y_0)$ in the direction of the unit vector $u=<u_1, u_2>$, denoted as $D_u f(x_0, y_0)$, is given by
$$D_u f(x_0, y_0) = ∇f \cdot u = |∇f||u| cos\theta = |∇f|cos\theta$$
where $\theta$ is the angle between the vectors $∇f$ and $u$.
Hence,
$1. D_u f(x_0, y_0) > 0$ whenever $|∇f|cos\theta > 0$
$2. D_u f(x_0, y_0) < 0$ whenever $|∇f|cos\theta < 0$
$3. D_u f(x_0, y_0) = 0$ whenever $|∇f|cos\theta = 0$
Note that in cases $1$ and $2$, the magnitude of the gradient is always positive. Hence, the sign of the directional derivative is determined by $\theta$. Thus, if we restrict $0 \le \theta \le \pi$,
$D_u f(x_0, y_0) =  |∇f|cos\theta > 0$ whenever $0 \le \theta < \frac{\pi}{2}$
$D_u f(x_0, y_0) =  |∇f|cos\theta < 0$ whenever $\frac{\pi}{2} < \theta \le \pi$
In case $3$, $D_u f(x_0, y_0) =  |∇f|cos\theta = 0$ whenever $\theta = \frac{\pi}{2}$ or $|∇f|=0$, the latter of which is true iff $f_x = f_y = 0$, meaning that $P_0$ is either an extremum or a saddle point.
Therefore, examine the angle $\theta$ between the gradient, or $∇f$, and the given vector ($v$ or $w$), in each graph, and use the criteria outlined immediately above to determine your answer.
If you're wondering where the gradient, or $∇f$, is on the graph, the gradient is a vector that is always perpendicular to a level curve (the proof of this is another matter entirely). It's not depicted in the graph, so you have to imagine it, which isn't hard. First, imagine a line tangent to the level curve that passes through your given point. Then, imagine a vector that is perpendicular to the tangent line whose initial point is the given point. Now, you will have two choices in terms of the direction of the gradient because there are two directions the gradient can point and still be perpendicular to the tangent line. For these purposes, I believe it is safe to assume that it will point in the direction in which the level curves are "increasing" because the direction of the gradient is the direction of maximum ascent (again, the proof of which is another matter). In other words, notice that your teacher has written a constant number next to each level curve... the gradient is perpendicular to the level curve and follows the direction in which those constant numbers are increasing.
