Is the projective curve $x^3+y^3=2z^3$ in $\mathbb P^2$ (defined over $\mathbb{C}$) singular or nonsingular? If singular, what are the types of these singularities?
For an affine curve, one would find the points at which the equation, and its first derivatives (with respect to both x and y) equal $0$. Would I do something similar for a projective curve? How should I go about determining if this curve is singular?
Given a homogeneous polynomial $f(x,y,z)\in \mathbb C[x,y,z]$ of degree $d\gt0$, the curve $V(f)\subset \mathbb P^2$
(= the zero-locus $\{f=0\}$) has as singularities the set of solutions $S \subset \mathbb P^2$ of the system $$\frac {\partial f}{\partial x}=\frac {\partial f}{\partial y}=\frac {\partial f}{\partial z}=0 \quad (\ast)$$.
Amazingly a point $s\in S$ automatically belongs to $V(f)$, i.e. $f(s)=0$, thanks to Euler's identity $$x\frac {\partial f}{\partial x}+y\frac {\partial f}{\partial y}+z\frac {\partial f}{\partial z }=df(x,y,z) \in \mathbb C[x,y,z]$$ In your case $f(x,y,z)=x^3+y^3-2z^3$, so that the system $(\ast)$ has as only solution $x=y=z=0$, which which does not correspond to any point in $\mathbb P^2$.
Hence $x^3+y^3-2z^3=0$ has no singularities: it is a smooth projective plane curve.
[Note that you don't have to do any calculations in the three canonical open affine subspaces $x=1,y=1, z=1$ of $\mathbb P^2$.]