Find the directions in which the directional derivative has the value 1 Can anyone show me how to adjust my work below so that it is a correct answer?  This is question number 14.6.28 in the 7th edition of Stewart Calculus.
Find the directions in which the directional derivative of $f(x,y)=ye^{-xy}$ at the point $(0,2)$ has the value 1.  
My work is:
$\nabla f(x,y) = <-y^2 e^{-xy}, e^{-xy}(1-xy)>$
$\nabla f(0,2) = <-4,1>$
$|\nabla f(0,2)| = \sqrt{17}$
$D_v f(x,y)=|\nabla f|\cos{\theta}=\sqrt{17}\cos{\theta}$
$\sqrt{17} \cos{\theta}=1$ when $\cos{\theta}={{\sqrt{17}}/17}$
$\theta =\arccos{{\sqrt{17}}/17}\approx +1.326$ and $-1.326$  

EDIT:
I tried the following, based on suggestions below:
$\vec{u}=<\cos{\theta},\sin{\theta}>$
$D_u f(x,y)=\nabla f(x,y)\cdot \vec{u}$
$D_u f(0,2)=\nabla f(0,2)\cdot \vec{u}=-4\cos{\theta}+\sin{\theta}=1$
I then plugged this into a spreadsheet and found that $-4\cos{\theta}+\sin{\theta}=1$ when $\theta = \pi/2 , 5\pi/2 , 9\pi/2 , ...$ and when $\theta = 4\pi/3 , 10\pi/3 , 16\pi/3 , ...$
Can anyone check the correctness of this approach?
Also, I found this result experimentally.  If it is correct, I would rather be able to find it using calculus.  
 A: The directional derivative is $\nabla f \bullet u$, where $u$ is a unit vector which points in the direction desired. What you want is the unit vector $u=(x,y)$; your del $f$ is $(-4,1)$ as you say, and then $\nabla f \bullet u$ is simply $-4x+1y$. Since it should be 1 you know that $-4x+y=1$, i.e. $y=1+4x$. Since $(x,y)$ is a unit vector you also know that $x^2+y^2=1$. 
So plugging in we have $x^2+(1+4x)^2=1$, which when you move the 1 over and expand gives the equation $17x^2+8x=0$. This factors as $x(17x+8)=0$. So either $x=0$ or else $x=-8/17$. Then plugging these into $y=1+4x$ gives the two unit vectors $(0,1)$ and $(-8/17,-15/17)$. 
[note the question said "find the directions" rather than "direction"; I think in general for a desired value of the directional derivative strictly between the gradient and the negative of the gradient, one usually has two directions. If you're on a hill not pointing straight up, and you find one way to walk so you're going up at a lesser rate than straight up, there should be another such direction...]
A: Let $u$ be the unit vector along which you would like to take the directional derivative of $f$. With $\theta$ the angle that $u$ makes with the $x$ axis,
$$
u = \left(\cos\theta,\sin\theta\right).
$$
You can check that $u$ is a unit vector.
Now, the directional derivative of $f$ in the direction of $u$, at the point $\left(0,2\right)$, is
$$
\nabla f \Big|_{\left(0,2\right)} \bullet u = \left(-4,1\right) \bullet \left(\cos\theta,\sin\theta\right) = -4 \cos \theta + \sin \theta.
$$
The problem states that this should be 1, so
$$
-4 \cos \theta + \sin \theta = 1.
$$
Now it remains to solve for $\theta$. You can probably do this with
$$
- \cos\left(a+b\right) = - \cos a \cos b + \sin a \sin b.
$$
