Directional Derivative Given the following equation:
$V(x,y,z)=5x^2-3xy+xyz$
Part 1: 
At point $P(3,4,5)$, find the rate of change in the direction of the vector $\langle1,1,-1\rangle$
Part 2: Find the direction in which $V$ changes most rapidly at $P(3,4,5)$
Part 3: Find the maximum rate of change at $P(3,4,5)$

I think I've managed to do part 1. Here's what I've done so far - but I am clueless as to how to find the max rate of change at $P$ and the direction in which the change occurs most rapidly.
$V_x(x,y,z)=10x+y(z-3)$
$V_y(x,y,z)=x(z-3)$
$V_z(x,y,z)=xy$
Based on the above partial derivatives,
$$\begin{align}
\nabla V(3,4,5) &=\left.\langle10x+y(z-3),\ x(z-3),\ xy\rangle\right|_{(3,4,5)} \\
&=\langle38,6,12\rangle
\end{align}$$
Unit vector of $\langle1,1,-1\rangle$ is $\vec u=\langle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\rangle$.
Directional derivative at $(3,4,5)$ in the direction of $\langle1,1,-1\rangle$ is:
$$\begin{align}
D_\vec uV(3,4,5) &= \nabla V(3,4,5) \cdot \vec u \\
&=\frac{32}{\sqrt{3}}
\end{align}$$
How should I proceed?
 A: In order to understand clearly what's going on, I'm going to restrict myself to two dimensions (i.e. $z=0$).
In this case
$$
\tilde V(x,y) = V(x,y,0) = 5 x^2 -3 xy
$$
The directional derivative in the direction $\vec{\alpha}$ is
$$
\frac{\partial \tilde V}{\partial \vec{\alpha}} = \nabla \tilde V \cdot \frac{\vec{\alpha}}{\|\vec{\alpha}\|}
$$
hence
$$
\frac{\partial \tilde V}{\partial \vec{\alpha}} = \frac{1}{\|\vec{\alpha}\|}\big(\tilde V_x \alpha_1 + \tilde V_y \alpha_2)
$$
where $\vec{\alpha} = (\alpha_1, \alpha_2)^T$.
As a function of $\alpha$, in $(3,4)$
$$
\frac{\partial}{\partial \vec{\alpha}} \tilde V(3,4) = \frac{18 \alpha_1 - 9 \alpha_2}{\sqrt{\alpha_1^2 + \alpha_2^2}}
$$
we
have
$$
\nabla_{(\alpha_1,\alpha_2)}\left(\frac{\partial}{\partial \vec{\alpha}} \tilde V(3,4)\right) = \begin{pmatrix} \frac{9\alpha_2(\alpha_1 + 2\alpha_2)}{(\alpha_1^2 + \alpha_2^2)^{3/2}}\\
-\frac{9\alpha_1(\alpha_1 + 2\alpha_2)}{(\alpha_1^2 + \alpha_2^2)^{3/2}}\\\end{pmatrix}
$$
and if $\alpha_1 = -2 \alpha_2$, $\nabla_{(\alpha_1,\alpha_2)} \left(\frac{\partial}{\partial \vec{\alpha}} \tilde V(3,4)\right) = 0$, hence, the direction in which $V$ changes critically is
$$
\vec{\alpha} = \frac{1}{\sqrt{5}}\begin{pmatrix}2 \\ -1 \end{pmatrix}
$$
$\hskip1.5in$
The maximum rate of change would be
$$
\frac{\partial}{\partial \vec{\alpha}} \tilde V(3,4) = 9\sqrt{5}
$$
in the direction $(2,-1)$.
Note that the contour $\tilde V(x,y) = 9$, which passes through $(3,4)$ can be expressed locally using the Implicit Function Theorem as a function $g(x)$, where
$$
g'(x) = -\frac{\tilde V_x}{\tilde V_y}
$$
as long as $\tilde V_y \neq 0$. In $(3,4)$
$$
g'(3) = 2
$$
The direction of min/max growth of a function is normal to the level curve passing trough it!
This is very usefull for finding local maxima/minima of a function, an there is a method called  Gradient Descent Method that depends upon this property.
A fantastic excercise would be to prove this result in general ;)
