Is Spivak wrong here, or am I just missing something? Chapter 1 Problem 18 has the reader doing various proofs with second-degree polynomial functions of the form $x^2 + bx + c$. My issue lies with problem 18d, but it uses knowledge from 18b and 18c, so bear with me.
Part B has the reader prove that 
if $b^2 - 4c < 0$ 
then 
$x^2 + bx + c > 0$ for all x
which was easy enough to show.
Part C has the reader prove that 
if x and y are both not 0, then $x^2 + xy + y^2 > 0$
which was accomplished by substituting $y = b$ and $y^2 = c$ to get 
$x^2 + bx + c$ 
and then following the same logic of showing that because $b^2 - 4c = y^2 - 4y^2 < 0$ then $x^2 + xy + y^2 > 0$.
Then we get to 18d, where the issue is. The question is:

18 (d) For which numbers $\alpha$ is it true that $x^2 + \alpha xy + y^2 > 0$, assuming x and y are both not 0?

from the answer book it says that $\alpha$ must satisfy $(\alpha y)^2 - 4y^2 < 0$ (taken from part B, by substituting  $b$ for $\alpha y$ and c for $y^2$)
$\Rightarrow$ $\alpha ^2 < 4$ $\Rightarrow |\alpha| < 2$
but this is clearly not true for many, many examples.
Say for instance that we let x = 2 and y = 3 in our equation $x^2 + \alpha xy + y^2$ then any positive $\alpha$ would make $x^2 + \alpha xy + y^2 > 0$ be true. $\alpha$ could equal 612 and the inequality would hold true in that situation.
I think the error in logic comes from Spivak's statement that 

$\alpha$ must satisfy $(\alpha y)^2 - 4y^2 < 0$

because it assumes that it's the only condition in which the polynomial could be positive, but it's clearly not the case. Also, it's easy to solve for the inequality because we can just do some basic manipulation to come up with the solution for $\alpha$
if

$x^2 + \alpha xy + y^2 > 0$

then 

$ \alpha xy > -(x^2 + y^2 )$

then

$\alpha > -(\frac{x^2 + y^2}{xy})$

In any case, let me know if there's any faults in my logic, or is there something I'm missing.
 A: Your logic is wrong. The problem asked for which $\alpha$'s such that: $x^2+\alpha xy + y^2 > 0$. You interpret it as: which $x,y$ such that there is an $\alpha$ such that : $x^2+\alpha xy + y^2 > 0$. These are non-equivalent statements. 
A: Let me explain.  The problem is 

18 (d) For which numbers α is it true that x2+αxy+y2>0, assuming x and y are both not 0?

The issue is that this sentence is not 100% precise. In fact, there are TWO ways to interpret it: 
(1). For which numbers α is it true that, for all x and y such the x$\neq$0 and y$\neq$0,  x2+αxy+y2>0?
This is the meaning intended by Spivak. 
The other way to interpret it is: 
(2). For all x and y such the x$\neq$0 and y$\neq$0, for which numbers α is it true that x2+αxy+y2>0? 
That is your interpretation of the problem. 
To be absolutely rigorous the sentence written in 18(d) in Spivak is ambiguous. HOWEVER, it is a common practice in Mathematics that sentences like the 18(d) mean (1) and do not mean (2). 
A: In the text book, the question is written as:

For which numbers $\alpha$ is it true that $x^2+\alpha xy+y^2 \gt 0$ whenever $x$ and $y$ are not both $0$? $\quad (\dagger_1)$

The proper interpretation of this question is the following:

Find an equivalent way of representing the following set: $S=\left\{\alpha \in \mathbb R\ |\  \forall x \forall y\ \big( y\neq 0 \text{ and } x \neq 0 \rightarrow x^2+\alpha xy+y^2 \gt 0 \big) \right\}$

In English: $\alpha \in S \iff $ for all $x$ and $y$ that are not simultaneously $0$, the inequality involving $\alpha$ holds. If $S$ were empty, then the following FOL statement would be false: $\exists \alpha \in \mathbb R: \forall x \forall y\ \big( y\neq 0 \text{ and } x \neq 0 \rightarrow x^2+\alpha xy+y^2 \gt 0 \big) $.
The goal of this problem is to properly characterize what set $S$ looks like...the answer, of course, is $S=(-2,2)$.

The other interpretation of $(\dagger_1)$, which is the incorrect interpretation, is:

Find an equivalent way of representing the following set: $T=\left\{\alpha \in \mathbb  R\ |\  \exists x \exists y \big(x\neq0 \text{ and } y \neq 0\text{ and } x^2+\alpha xy+y^2 \gt 0\big)\right\}$

In your 'counter example', you found that when $x=2$ and $y=3$, any $\alpha \in \mathbb R$ would satisfy the inequality. In this case, $T=\mathbb R$.
