I appreciate your time. If anyone can explain this problem, I would be most grateful. I need to understand this for a test, but I was not given any explanation.
Assume that $ ∇f(x,y) ≠ 0 $. Show that the value of $ {D_{\mathbf{u}} f}(x,y) $ is
largest when $ \mathbf{u} = |∇f(x,y)| \cdot ∇f(x,y) $;
smallest when $ \mathbf{u} = − |∇f(x,y)| \cdot ∇f(x,y) $.
We have $\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial x})$. Let $u = (u_1, u_2)$. Then we have
$$\nabla_u f (x, y)= \frac{d}{dt} f(x+ tu_1, y+tu_2)|_{t=0} = \frac{\partial f}{\partial x}u_1 + \frac{\partial f}{\partial x}u_2 = \langle u, \nabla f (x, y)\rangle\ .$$
The second equality follows from chain rule. To maximize the last quantity, note
$$ \langle u, \nabla f (x, y)\rangle =|u| |\nabla f(x, y)| \cos \theta\ ,$$
where $\theta $ is the angle between the two vectors. Then in order to maximize this term, we should choose $u$ such that $\theta=0$. That is, $u$ is pointing to the same direction to $\nabla f(x, y)$. Thus
$$u =\frac{\nabla f(x, y)}{|\nabla f(x, y)|}$$
if you insist that $u$ has length one. Similar argument works for the minimum.