Flex coordinates for a smooth plane cubic ($F=a{X}^3+Z(b{Y}^2+cXY+d{X}^2)+Z^2(eY+fX)+g{Z}^3$)

Put a flex at $[0:1:0]$ with tangent line $Z=0$. The tangent section has a triple zero there, so $F(X,Y,0)=a{X}^3$ with $a\ne 0$. Smoothness gives $b\ne 0$. In characteristic different from two, replacing $Y$ by $Y+(cX+eZ)/(2b)$ completes the square and produces a Weierstrass equation of an elliptic curve using only projective linear coordinate changes.