I've asked about this already here (thanks to user10354138 for the quick and good answer!) but when I showed the solution to my instructor, he said I should follow first his procedure for getting the dual curve and see if both methodologies yield the same result.
My instructor's procedure:
Consider a projective elliptic curve $E$ given by $$F(x,y,z)=y^2z-x^3$$ Then get the partial derivatives $$\frac{\partial F}{\partial x}=-3x^2$$ $$\frac{\partial F}{\partial y}=2yz$$ $$\frac{\partial F}{\partial z}=y^2$$
Now, define the mapping
\begin{align*} \phi: (x:y:z)&\mapsto(\frac{\partial F}{\partial x}:\frac{\partial F}{\partial y}:\frac{\partial F}{\partial z})\\ &\mapsto(-3x^2:2yz:y^2) \end{align*}
Let $$X=-3x^2$$ $$Y=-2yz$$ $$Z=-y^2$$
Now, we can do two things here: first is to get $x,y,z$ in terms of $X,Y,Z$ and substitute to $E$ to get the dual curve, and second is to get the terms of $E$ and represent it in terms of $X,Y,Z$.
First method:
$$x=(-\frac{X}{3})^\frac{1}{2}$$ $$y=(-Z)^\frac{1}{2}$$ $$z=-\frac{Y}{2y}=-\frac{Y}{2(-Z)^\frac{1}{2}}$$
Then, substituting these to $E$,
$$\frac{ZY}{2(-Z)^\frac{1}{2}}-(-\frac{X}{3})^\frac{3}{2}=0$$ $$\frac{Z^2Y^2}{-4Z}=(-\frac{X}{3})^3$$ $$\frac{ZY^2}{-4}=-\frac{X^3}{27}$$ $$27ZY^2=4X^3$$
This is the dual curve of $E$.
Second method:
$$y^2z=(-\frac{Y}{2})(-Z)^\frac{1}{2}$$ $$-x^3=-(-\frac{X}{3})^\frac{3}{2}$$
Adding these together,
$$(-\frac{Y}{2})(-Z)^\frac{1}{2}-(-\frac{X}{3})^\frac{3}{2}=0$$ $$(\frac{Y^2}{4})(-Z)=(-\frac{X}{3})^3$$ $$-\frac{ZY^2}{4}=-\frac{X^3}{27}$$ $$\frac{ZY^2}{-4}=-\frac{X^3}{27}$$ $$27ZY^2=4X^3$$
This is the dual curve of $E$.
Now, how can I do this for projective elliptic curves in general (given in Weierstrass form $y^2z=x^3+axz^2+bz^3$)? I'm particularly stuck in getting the equation of dual curve after defining the mapping $\phi$.