I'm reading Singularities of Differential Maps by Arnold, Gusein-Zade, and Varchenko, and I'm a bit confused about their definition of a degenerate critical point. Unlike what I've found on the internet, they use two single variable equations, $y=x^2$ and $y=x^3$. In such a case I don't think it would apply to use the Hessian. They then define that "a critical point of a smooth function is said to be non-degenerate if the second differential of the function at that point is a non-degenerate quadratic form".
If I take the derivative twice, $y=x^2$ becomes $y''=2$ which doesn't have a zero so I'm guessing that it would be non-degenerate? However, if I take the derivative of $y=x^3$ twice, it becomes $y''=6x$, which has a zero only at $x=0$, which would seem to make it non-degenerate as well. However, the book says that $y=x^3$ is degenerate. What am I missing here?
$x^3$ is degenerate, because the second derivative at $0$ is $0$. $x^2$ is not degenerate, because the second derivative at $0$ is $2$.
In the language of differentials, the second differential of $x^3$ at $x=0$ is $0 \; dx\; dx$ and the second differential of $x^2$ at $x=0$ is $2 \; dx \; dx$.