The question states:
Evaluate the directional derivative of $f(x,y,z)=x^2+2y^2+3z^2$ at $(1,1,0)$ in the direction of $\vec i - \vec j +2\vec k$.
I learned that $f'(\vec a; \vec y) = \nabla f(\vec a)\cdot \vec y$. Now $\nabla f(\vec a)$ is the gradient vector, which has components $(D_1f(\vec a), ..., D_n f(\vec a))$.
Therefore I first calculated the gradient vector by calculating each partial derivative, and I obtained $\nabla f(\vec x) = (2x, 4y, 6x)$. I then substituted in $(1,1,0)$ and obtained $\nabla f(1,1,0) = (2, 4, 0)$. Finally, taking the dot product with $(1,-1,2)$, I obtained $f'((1,1,0); (1,-1, 2)) = -2$. However, the answer I am given is $-2/\sqrt 6$. What am I doing wrong? The theorem I am given, with proof, states: Assume $f$ is differentiable at $\vec a$ with total derivative $T_a$. Then the derivative $f'(\vec a; \vec y)$ exists for every $\vec y$ in $R^n$ and we have $$T_a(\vec y) = f'(\vec a; \vec y)$$. Moreover, $f'(\vec a; \vec y)$ is a linear combination of the components of $y$. In fact, if $y = (y_1, ..., y_n$, we have $$f'(\vec a; \vec y)=\sum^n_{k=1}D_kf(\vec a)y_k$$. This can be written as a dot product: $\nabla f(\vec a)\cdot y$.
The proof is as follows: The first formula holds trivially if $\vec y = \vec O$ since both $T_a(\vec O) = 0$ and $f'\vec a; \vec O = 0$.Therefore we can assume that $\vec y \neq \vec O$. Since $f$ is differentiable at $\vec a$ we have a Taylor formula, $$f(\vec a + \vec v) = f(\vec a) + T_a(\vec v) + ||\vec v||E(\vec a, \vec v)$$ for $||\vec v|| < r$ for some $r>0$, where $E(\vec a, \vec v)\rightarrow 0$ as $||\vec v|| \rightarrow 0$. In this formula we take $\vec v = h\vec y$, where $h\neq 0$ and $|h|~ ||\vec y||<r$. Then $||\vec v|| < r$. Since $T_a$ is linear we have $T_a(\vec v) = T_a(h\vec y) = hT_a(\vec y)$. Therefore the above Taylor formula gives us $$\frac{f(\vec a + h\vec y)-f(\vec a)}{h}=T_a(\vec y) + \frac{|h| ~ ||\vec y||}{h}E(\vec a, \vec v).$$ Since $||\vec v|| \rightarrow 0$ as $h \rightarrow 0$ and since $|h|/h = ±1$, the right-hand member of the above equation tends to the same limit $T_a(\vec y)$ as $h\rightarrow0$. Therefore the left-hand member tends to the same limit. This proves the theorem's first formula.
Now we use the linearity of $T_a$ to deduce the second formula. If $\vec y = (y_1,...,y_n)$ we have $\vec y = \sum^n_{k=1}y_k\vec{e_k}$, hence $$T_a(\vec y) = T_a(\sum^n_{k=1}y_k\vec{e_k})=\sum^n_{k=1}y_kT_a(\vec {e_k})=\sum^n_{k=1}y_kf'(\vec a; \vec {e_k})=\sum^n_{k=1}y_kD_kf(\vec a).$$
Your formula for the directional derivative is wrong. It is correct if $ y$ is a unit vector. In general you have to divide by the norm of the vector which is $\sqrt 6$ in this case.
It is worth observing that the directional derivative cannot depend on the length of the vector; it can only depend on the direction of the vector.
EDIT: there are conflicting definitions of directional derivatives (as I just discovered) and whatever you have done is right with your definition. The answer provided is based on a different definition.