Hello, i_zz_y_ill!
I assume you know to set up Lagrnge multipliers.
Find the local max and minima from origin to curve $\displaystyle x^3+y^36xy\:=\:0$
Answer are: minimum (0, 0), maximum (3,3)
We want to minimize the distance from point $\displaystyle (x,y)$ to the origin.
The distance function is: .$\displaystyle d \:=\:\sqrt{x^2+y^2}$
For convenience, we can minimize the square of that distance: .$\displaystyle D \:=\:x^2+y^2$
Our function is: .$\displaystyle f(x,y,\lambda) \;=\;(x^2+y^2) + \lambda(x^3 + y^3  6xy)$
Set the partial derivatives equal to 0, and solve the system . . .
. . $\displaystyle \begin{array}{cccccc}f_x &=& 2x + 3\lambda x^2  6y &=& 0 &[1] \\
f_y &=& 2y + 3\lambda y^2  6x &=& 0 & [2] \\
f_{\lambda} &=& x^3 + y^3  6xy &=& 0 & [3] \end{array}$
$\displaystyle \begin{array}{ccccc}\text{From [1], we have:} & \lambda &=& \dfrac{6y2x}{3x^2} & [4] \\ \\[3mm]
\text{From [2], we have:} & \lambda &=& \dfrac{6x2y}{3y^2} & [5] \end{array}$
Equate [4] and [5]: .$\displaystyle \frac{6y2x}{3x^2} \:=\:\frac{6x2y}{3y^2} \quad\Rightarrow\quad 3x^3  3y^3  x^2y + xy^2 \:=\:0$
Factor: .$\displaystyle 3(x^3y^3)  xy(xy) \:=\:0 \quad\Rightarrow\quad 3(xy)(x^2+xy+y^2)  xy(xy) \:=\:0$
Factor: .$\displaystyle (xy)\bigg[3(x^2+xy+y^2)  xy\bigg] \:=\:0 \quad\Rightarrow\quad 3(xy)(3x^2 + 2xy + 3y^2) \:=\:0$
From $\displaystyle xy \:=\:0$, we have: .$\displaystyle \boxed{y \:=\:x}\;\;[6]$
And $\displaystyle 3x^2 + 2xy + 3y^2 \:=\:0$ has no real factors.
Substitute [6] into [3]: .$\displaystyle x^3 + x^3  6x^2 \:=\:0 \quad\Rightarrow\quad 2x^3  6x^2 \:=\:0 \quad\Rightarrow\quad 2x^2(x3) \:=\:0$
. . Hence, we have: .$\displaystyle x \:=\:0,3 \quad\hdots\quad y \:=\:0,3$
Therefore, the critical points are: .$\displaystyle (0,0),\;(3,3)$
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I could be wrong, but I think this is the Folium of Descartes.
I has a leafshaped loop in Quadrant 1
. . . $\displaystyle \text{and approaches }x + y + 3\:=\:0 \text{ as an asymptote.}$
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