Let $f(x,y,z) = x^3 + y^3 + z^3$. The dual curve of $f$ is defined to be the curve whose coordinates are given by $$\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right).$$
Evaluating these partial derivatives is simple, but determining the curve has been some issue. Does anyone have any nice way of determining dual curves and can help me with this example?
For reference: https://en.wikipedia.org/wiki/Dual_curve
Let $(p,q,r)$ be a point on the given surface. To get an implicit equation for the dual surface, we need to eliminate $p,q,r$ from the following equations: $$u-\lambda \frac{\partial f}{\partial x}(p,q,r) = u- 3\lambda p^2 =0\tag{1}$$ $$v-\lambda \frac{\partial f}{\partial y}(p,q,r) = v- 3\lambda q^2 =0\tag{2}$$ $$w-\lambda \frac{\partial f}{\partial z}(p,q,r) = w- 3\lambda r^2 =0\tag{3}$$ $$up+vq+zr=0 \tag{4}$$ Solving $(4)$ for $p=-\dfrac{qv+rw}{u}$ and substituting it in $(1)$ yields $$3\lambda q^2v^2+6\lambda qrvw+3\lambda r^2w^2-u^3=0 \tag{5}$$ The first term of $(5)$, $3\lambda q^2$ can be replaced by $v^3$ following $(2)$. Thus, solving for $q=\dfrac{u^3-3\lambda r^2 -v^3}{6\lambda r v w}$ and substituting it back in $(2)$ gives: $$9 \lambda^2 r^4 w^4-6 \lambda r^2 u^3 w^2-6 \lambda r^2 v^3 w^2+u^6-2u^3v^3+v^6=0 \tag{6}$$ $(6)$ contains the term $r$ only in powers of 2 and 4 which can be replaced using $(3)$ with $r^2=\dfrac{w}{3 \lambda}$. Thus the equation of the dual surface after eliminating $p,q,r$ is $$u^6-2u^3v^3-2u^3w^3+v^6-2v^3w^3+w^6=0$$
In the following figure, the red surface is the dual surface of the blue surface $x^3+y^3+z^3=0$