I have a question about the dual curve to the curve $C$ cut out by the equation $F(X,Y,Z) = XY^2 - Z^3 = 0$ in $\mathbb{P}^2$. (Assume that everything is over an algebraically closed field of characteristic zero.)
The set of tangent lines to this curve is given by the points \begin{equation} \left[\frac{\partial F}{\partial X}:\frac{\partial F}{\partial Y}:\frac{\partial F}{\partial Z}\right] = [Y^2:2XY:-3Z^2] \end{equation} in the dual space $(\mathbb{P}^2)^*$.
Of course, we'll have to throw out the cusp $[1:0:0]$. But away from this cusp, $Y \neq 0$, so the tangent lines are the points \begin{equation} [Y^2:2XY:-3Z^2] = [Y^3:2XY^2:-3YZ^2] = [Y^3:2Z^3:-3YZ^2] \end{equation}
So by inspection the dual curve is $27UV^2 + 4W^3 = 0$ in the coordinates $[U:V:W]$ on $(\mathbb{P}^2)^*$.
This has degree $3$. But the theory asserts that the degree of the dual curve to a curve of degree $d$ is $d(d-1)$ (at least for smooth curves), or in this case, $3 \cdot 2 = 6$.
Where did the other $3$ degrees go? Does this have to do with the fact that the curve has a cusp, and what should I do to recover the equation of the asserted degree $6$ dual curve?
EDIT: OK, I did some reading and turns out that I need to add the tangents at the cusp back in. If we define to dual curve to be the set of all lines that intersect the curve at a point with multiplicity $>1$, then every point passing through the cusp $[1:0:0]$ is a tangent line, so I need to add in the line in $(\mathbb{P}^2)^*$ of lines passing through $[1:0:0]$, i.e., $U = 0$. In fact, since this is a cusp, I need to add in the cube of this (this probably has something to do with the fact that the "maximal" intersection multiplicity at this point by some line, in this case the line $Y = 0$, is $3$), so the dual curve should be something like \begin{equation} U^3(27UV^2+4W^3) = 0. \end{equation}
This is a degree 6 equation, so all seems to be good.
Can somebody confirm this? It would be great also if somebody could explain why cusps mean I should add in the cube of the line (and why nodes mean the square of the line).
If $C$ is a non-singular curve of degree $d$ then the degree of its dual curve $C^*$ is indeed $d^*=d(d-1)$.
However this does not hold for singular curves. Plücker formula gives the degree of $C^*$ in terms of the degree and singularities of $C$. The formula is $d^*=d(d-1)-2\delta-3\kappa$, where $\delta$ is the number of ordinary nodes and $\kappa$ is the number of ordinary cusps, higher order singularities are counted as multiple ordinary singularities.
Back to the question, as $C$ is cut out by the equation $XY^2−Z^3=0$ it has degree $3$, no nodes and one ordinary cusp. By the Plücker formula the degree of $C^*$ is $d^*=3\cdot2 -3=3$ exactly as you found in the first place.
Lets say a bit more about why the singularities appear in the Plücker formula.
Note that even for a smooth curve $C$ its dual $C^*$ is not in general smooth. For instance, if $C$ has a bitangent i.e. a line that is tangent at two distinct points. Then this line corresponds to an ordinary node in $C^*$. More generally, if the line $L$ is tangent to $C$ at $r$ distinct points none of which are inflection points then its dual is a point of $C^*$ with $r$ distinct tangent directions.
If $P$ is an inflection point of $C$ (the intersection multiplicity of the tangent and $C$ at $P$ is at least $3$) then the dual of the tangent is a cusp.
Furthermore if our algebraically closed ground field is also of characteristic $0$ then $(C^*)^*=C$ and we obtain a correspondence cusp$\leftrightarrow$inflection-point and node$\leftrightarrow$bitangent.