Does this function define first-order ordinary differential equations? I've read 3 different books but, this condition seems to fit them all:

It's homogeneous if:
  $y'=f(tx,ty)=f(x,y)$

Is this correct?
 A: This is one sense in which a differential equation can be homogeneous, but be cautioned that this is not nearly the more common sense, which is very different!

The usual sense of homogeneous in the context of differential equations is the following: A linear ordinary differential equation is one of the form
$$\frac{d^m}{dx^m} y(x) + A_{m - 1}(x) \frac{d^{m - 1}}{dx^{m - 1}} + \cdots + A_1(x) \frac{d}{dx} y(x) + A_0(x) y = F(x).$$
It is furthermore homogeneous iff $F(x)$ (called the source term) is $0$, or equivalently, iff $y(x) = 0$ is a solution. A key feature of homogeneous linear o.d.e.s is that their solutions comprise a vector space under the usual addition and scalar multiplication of functions.
We see immediately that an o.d.e. $y' = f(x, y)$ is homogeneous if $f(x, y)$ is linear in $y$, that is, if $\frac{\partial^2 f}{\partial y^2} = 0$.

Wikipedia describes a second notion of homogeneous first-order equations that I hadn't encountered before. It's closer to the notion you describe but still distinct. A first-order differential equation of the form $$P(x, y) \,dx + Q(x, y) \,dy = 0$$ is of homogeneous type if $P, Q$ are homogeneous functions of the same degree: A function $R(x, y)$ is homogeneous of degree $m$ iff $R(\lambda x, \lambda y) = \lambda^m R(x, y)$. Rearranging, we can rewrite such an equation (ignoring the places where $Q$ takes on the value zero) as $$\frac{d}{dx} y(x) = -\frac{P(x, y)}{Q(x, y)},$$ and the right-hand side satisfies $$-\frac{P(\lambda x, \lambda y)}{Q(\lambda x, \lambda y)} = -\frac{\lambda^m P(x, y)}{\lambda^m Q(x, y)} = -\frac{P(x, y)}{Q(x, y)},$$ so any homogeneous first-order differential equation of homogeneous type can be put in the indicated form.
Conversely, we can rewrite the given equation $y' = f(x, y)$ as $$-f(x, y) \,dx + dy = 0,$$ and both of the coefficients are homogeneous of degree zero, hence so view the given equation as one of homogeneous type, and hence the notion in the question agrees with the notion of homogeneous type, at least up the zero set of $Q$.
