Show that the point $[0,1,0]$ is a non-singular point of $C$

Question

I am self-studying Rational Points over Elliptic Curves (Silverman & Tate) and am working on the following question:

$1.17$ Let $C$ be a cubic curve in the projective plane given by the homogeneous equation $$Y^2Z = X^3 + aX^2Z + bXZ^2 + cZ^3$$ Verify that the point $[0,1,0]$ at infinity is a non-singular point of $C$.

I am hoping to get some feedback on my proof, and if it is wrong, could someone point me in the right direction or give a correct proof? Here's what I have:

Let $F(X, Y, Z) = X^3 + aX^2Z + bXZ^2 + cZ^3 - Y^2Z$. Then we have that \begin{align} F_X &= 3X^2 + 2aXZ + bZ^2 \\ F_Y &= 2YZ \\ F_ Z &= aX^2 + 2bXZ + 3cZ^2 - Y^2. \end{align} So at the point $[0,1,0]$, we have that \begin{align} F_X (0) &= 0 \\ F_Y (0) &= 0 \\ F_Z (0) &= 1, \end{align} so the point is a non-singular point.

Is this all I have to do?

Answer 1

You have misinterprted the question. You are not being asked to hunt for singular points on the curve. You are required to show that the particular point (0:1:0) is a non-singular point, so what is required is to show that, regardless of what values are taken by the parameters $a,b,c$, all three partial derivatives cannot be 0 at (0:1:0).