I am trying to prove the following claim:
Let $K$ be a field and let $f \in K[x]$ be a monic cubic polynomial. Define the cubic curve $C: y^2 = f(x)$. If $C$ has a singularity $(a, b)$, then $b = 0$ and $a$ is a solution to $f(x) = 0$.
I feel like I have a proof, but I'm not sure if it is correct because I feel that it is too simple.
Proof. Write $f(x) = x^3 + ax^2 + bx + c$ where $a, b, c \in K$ and consider the polynomial $P(x, y) = y^2 - f(x)$ which defines the curve $C$. Suppose $(p, q)$ is a singularity of $C$. Then this implies that $$ \frac{\partial P}{\partial y}(p, q) = 2q = 2 \sqrt{f(p)} = 0. $$ It thus follows that $q = 0$ and $f(p) = 0$. $\, \, \blacksquare$
Would you agree that this is a proof? If not, can you point out where the error is? Thanks in advance.