Let $f(x, y) = y^2 - g(x) \in \mathbb{R}[x, y]$, where $g \in \mathbb{R}[x]$ is a monic polynomial of degree $3$. Let $C = V_f(\mathbb{R})$ and $(0, 0)$ a point on $C$.
I have to prove that $g$ is of the form $g(x) = x^2(x-a), a \in \mathbb{R}$ if and only if $(0, 0)$ is a singular point of $C$.
I have shown that, if $g(x) = x^2(x-a)$, then $(0, 0)$ is a singular point of $C$. I tried the other direction, but I don´t know how to continue. Here is my attempt :
Assume that $(0, 0)$ is a singular point of $C$. Then we have $f(0, 0) = -g(0) = 0$, $\frac{\partial f}{\partial y}(0, 0) = 0$ and $\frac{\partial f}{\partial x}(0, 0) = -\frac{\partial g}{\partial x}(0) = 0$. It follows that $g(0) = \frac{\partial g}{\partial x}(0)$.
How can I continue and conclude that $g$ is of the form $g(x) = x^2(x-a)$ ?
Thanks for your help.
$g$ is of the form $x^3 + ax^2 + bx + c$.
$g(0) = 0$ implies $c = 0$ so write $g$ as $x(x^2 + ax + b)$. Now,$g'(0) = 0$ will imply that $b = 0$, so write $g$ as $x^2(x + a)$.