In the preface of Foundations of Differential Calculus there's a section that says:
Thus, if the quantity $x$ is given an increment $\omega$, so that it becomes $x + \omega$, its square $x^2$ becomes $x^2 + 2x\omega + \omega^2$, and it takes the increment $2x\omega + \omega^2$. Hence, the increment of $x$ itself, which is $\omega$, has the ratio to the increment of the square, which is $2x\omega+\omega^2$, as $1$ to $2x+\omega$. This ratio reduces to $1$ to $2x$, at least when $\omega$ vanishes.
$f(x) = x^2$
$f(x+w) = (x+\omega)^2 = x^2 + 2x\omega + \omega^2$
$(x+\omega)^2 - x^2 = 2x\omega+\omega^2$
And then the part I don't understand where $\omega = 2x\omega+\omega^2$ goes to "as $1$ to $2x+\omega$".