I don’t recall where I was going with these equations but just now I think that projective transformations are the way to find canonical curves.

The general third degree curve in the Euclidean plane has the equation:
φ = ax3 + bx2y + cxy2 + dy3 + ex2 + fxy + gy2 + hx + iy + j = 0.
If we multiply each coefficient by some non zero field element, we define the same curve. Suppose we translate the origin: (x, y) = (x' + α, y' + β):
φ = a(x' + α)3 + b(x' + α)2(y' + β) + c(x' + α)(y' + β)2 + d(y' + β)3 + e(x' + α)2 + f(x' + α)(y' + β) + g(y' + β)2 + h(x' + α) + i(y' + β) + j
= ax'3 + bx'2y' + cx'y'2 + dy'3 + (3aα+e)x'2 + (2bα+2cβ+f)x'y' + (3dβ+g)y'2 + (3aα2+2bαβ+cβ2+h)x' + (bα2+2cαβ+3dβ2+i)y'
+ (aα3+bα2β+cαβ2+dβ3+eα2+fαβ+gβ2+hα+iβ+j)
= ax'3 + 3ax'2α + 3ax'α2 + aα3


our 3rd order terms remain unchanged, out 2nd order terms are
If we translate the origin to some point on the curve, j becomes 0.