<s>[INST] Find the points on the curve $x^4+y^4+3xy=2$ closest and farthest to the origin I need to find the points on the curve $x^4+y^4+3xy=2$ that are closest and farthest to the origin. 
I believe this might be a Lagrange multiplier problem, but I am not sure. I was thinking that maybe minimizing/maximizing the function would be the way to go.
 [/INST] I saw people recommending to use lagrange multiplier, but I know none of that. Instead, I tried this.
Firstly, you want to maximise/minimise (sqrt) $x^2+y^2$ with $x^4+y^4+3xy-2=0$. Consider $x^2+y^2=t$ and the intersection point $(u, v)$. i.e.
$$u^2+v^2=t, u^4+v^4+3uv-2=0$$
If at the point $(u, v)$, we can get both of the $\frac{dy}{dx}$ to be equal, then it means the curve $x^2+y^2-t=0$ and $x^4+y^4+3xy-2=0$ are tangent to each other.
Finding $\frac{dy}{dx}$ of both functions
Implicit differentiation gives
$$x^2+y^2=t \implies \frac{dy}{dx}=-\frac{x}{y}$$
$$x^4+y^4+3xy-2=0 \implies \frac{dy}{dx}=-\frac{4x^3+3y}{4y^3+3x}$$
We want to equal those. After simplifying
$$4xy^3+3x^2=4x^3y+3y^2$$
$$3(x^2-y^2)=4xy(x^2-y^2)$$
$$xy=\frac{3}{4} \textrm{ or } x=\pm y$$
If $xy=\frac{3}{4}$, $x^4+y^4+3(\frac{3}{4})-2=0 \implies x^4+y^4=-\frac{1}{4}$ which is impossible since LHS is non-negative.
Therefore, $x=\pm y$
**If ** $x=y$, $2x^4+3x^2-2=0\implies(2x^2-1)(x^2+2)=0\implies x=\pm \frac{1}{\sqrt{2}}, y=\pm \frac{1}{\sqrt{2}}, \sqrt{x^2+y^2}=1$
**If ** $x=-y$, $2x^4-3x^2-2=0\implies(2x^2+1)(x^2-2)=0\implies x=\pm \sqrt{2}, y=\mp \sqrt{2}, \sqrt{x^2+y^2}=2$
$$\therefore 1<\sqrt{x^2+y^2}<2$$
Side note: To the people doing lagrange multiplier, stop and calm down :) As mentioned in the comments, (apparently) this is the foundation of lagrange multiplier - Langrange multiplier is basically a generalisation of this to any constraint-equations and more variables :)
</s>