Since an elliptic curve given by the homogenized polynomial $$y^2z=x^3+axz^2+bz^3$$ is a plane projective curve, we can get its dual.
From this Wikipedia link, eliminating $p$, $q$, $r$, and $λ$ from the following equations, $$X-\lambda \frac{\partial f}{\partial x}(p,q,r) = X- \lambda (-3p^2-ar^2) =0\tag{1}$$ $$Y-\lambda \frac{\partial f}{\partial y}(p,q,r) = Y- \lambda (2qr) =0\tag{2}$$ $$Z-\lambda \frac{\partial f}{\partial z}(p,q,r) = Z- \lambda (q^2-2apr-3br^2) =0\tag{3}$$ $$Xp+Yq+Zr=0\tag{4}$$ yields the equation of the dual curve. However, no matter how I try to solve it, I always fail to eliminate all those variables.
Any help will be greatly appreciated.
I'll warn you in advance that this is not going to be pretty.
The $\lambda$ obviously doesn't matter because we are working with projective coordinates $X:Y:Z$. So we might as well put $\lambda=1$. The final equation just asserts $p:q:r$ is on the original curve $q^2r-p^3-apr^2-br^3=0$ when you substitute in $X,Y,Z$, but let's keep it in the original form $Xp+Yq+Zr=0$.
First step is to use $Xp+Yq+Zr=0$ to eliminate one of $p,q,r$. So let's try eliminating $p$, giving \begin{align*} X^3+3(qY+rZ)^2+ar^2X^2&=0\\ Y-2qr&=0\\ ZX-q^2X-2ar(qY+rZ)+3br^2X&=0 \end{align*} Next, eliminate $q$ using $Y=2qr$, \begin{align*} 0&=r^2X^3+3(Y^2/2+r^2Z)^2+ar^4X^2\\ &=r^4(aX^2+3Z^2)+r^2(X^3+3Y^2Z)+\frac34Y^4 \\ 0&=r^2ZX-\frac14XY^2-ar^2(Y^2+2r^2Z)+3br^4X\\ &=r^4(3bX-2aZ)+r^2(ZX-aY^2)-\frac14XY^2 \end{align*} Finally, eliminate $r^2$: \begin{align*} 0&=\operatorname{Res}_{r^2}(r^4(aX^2+3Z^2)+r^2(X^3+3Y^2Z)+\frac34Y^4,r^4(3bX-2aZ)+r^2(ZX-aY^2)-\frac14XY^2)\\ &=\det\begin{pmatrix} aX^2+3Z^2&X^3+3Y^2Z&\frac34Y^4\\ &aX^2+3Z^2&X^3+3Y^2Z&\frac34Y^4\\ 3bX-2aZ&ZX-aY^2&-\frac14XY^2\\ &3bX-2aZ&ZX-aY^2&-\frac14XY^2 \end{pmatrix} \end{align*} giving $$ \frac3{16} X^2 Y^2 (4 a^3 Y^6 - a^2 X^4 Y^2 - 24 a^2 X Y^4 Z + 18 a b X^2 Y^4 + 4 a X^5 Z + 30 a X^2 Y^2 Z^2 + 27 b^2 Y^6 - 4 b X^6 - 36 b X^3 Y^2 Z - 54 b Y^4 Z^2 + 4 X^3 Z^3 + 27 Y^2 Z^4)=0 $$ The factor $X^2Y^2$ was introduced by our desire to keep everything as polynomial when eliminating $p,q$ (we have multiplied equation by $X^2$, $r^2$), and not actually part of the dual curve. So the end result is $$ 4 a^3 Y^6 - a^2 X^4 Y^2 - 24 a^2 X Y^4 Z + 18 a b X^2 Y^4 + 4 a X^5 Z + 30 a X^2 Y^2 Z^2 + 27 b^2 Y^6 - 4 b X^6 - 36 b X^3 Y^2 Z - 54 b Y^4 Z^2 + 4 X^3 Z^3 + 27 Y^2 Z^4 = 0. $$