Irreducible cubic curve in normal form I have been given a cubic curve $F : X^3 - XZ^2 + YZ^2 = 0$ in the projective plane. I have that a singular point of the curve is $[0,1,0]$. I am asked to find a change of variable which puts $F$ into 'normal form', i.e $Y^2Z - G(X,Z)$ where $G(X,Z) = X^3 + bX^2Z + cXZ^2 + dZ^3$ for $b,c,d \in \mathbb{C}$.
I have been trying to get this for quite a while now and can't seem to get the change correct. I seem to be stuck with the two $Z^2$ terms in $F$.
I have a method but that relies on the point being a flex, which clearly isn't the case here. My notes indicate that it is common to just go with "guess work"... Is there a better method I could be using or is this just down to practice (and if so does any one know any good resources or problem sources for this)? Many thanks!

There is an "algorithm" for transforming cubic curves to the (long) Weierstrass form, and from that to the short Weierstrass form, which is usually introduced when studying elliptic curves. You can find it in its various forms here or here. But that involves a lot of tedious work.

Here is how I'd solve it manually: Usually it is helpful to derive the transformation step by step:
First note that $X^3 -XZ^2+YZ^2 = X^3 - (X-Y)Z^2$. By looking at the degrees we can easily see that whatever variables in the linear combination that we replace $X$ with will appear as third degree terms, and similarly what we plug in $Z$ will appear as only second degree terms. This suggests that $Z$ will be our future $Y$, so let us apply $Z \mapsto Y$ and $Y \mapsto Z$. We get
$$X^3 - (X-Z)Y^2$$
So now we want to get rid of  the $X$ as a coefficient of $Y^2$ as the term with $Y^2$ should not include a factor $X$. We can do this by replacing $Z$ with $X+Z$ so this yields
$$Y^2Z + X^3$$
Now we are almost there, we need to change the sign of the $X^3$ term which helpfully has and odd degree so we can just replace $X$ by $-X$ and get
$$Y^2Z -X^3$$

The important thing we need to check is whether our compound linear transformation is actually invertible. If we write the substitution as a matrix vector equation
$$\begin{bmatrix} X \\ Y  \\ Z \end{bmatrix} = \begin{bmatrix} * & * & * \\  * & * & * \\  * & * & *  \end{bmatrix}\begin{bmatrix} X' \\ Y'  \\ Z' \end{bmatrix}$$
this is equivalent to this matrix being invertible.  If you recall from linear algebra: The modulus of the determinant does not change if we swap rows, or add the multiple of one one row to another. This corresponds to swapping variables and replacing one variable by itself plus a multiple of another. And this is all we did above so this is indeed an invertible transformation. Alternatively you can also go through each step and actually determine this transformation matrix. 
Lastly: If you want to play around with that without too much writing use some CAS, I used Maxima, and you can try my "program" online: Try it online!
