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.

By Lagrange, we minimize
$$x^2+y^2+\lambda(x^4+y^4+3xy-2).$$
The derivatives on $x$ and $y$ yield
$$2x+\lambda(4x^3+3y)=2y+\lambda(4y^3+3x)=0,$$ and after elimination of $\lambda$,
$$2x(4y^3+3x)=2y(4x^3+3y)$$ or 
$$(8xy-6)(y^2-x^2)=0.$$
The solutions $y=\pm x$ imply
$$2x^4\pm3x^2-2=0,$$ giving
$$x^2+y^2=2x^2=\frac{\mp3\pm5}2$$ and only the positive solutions $1$ and $4$ are valid.
Next, with $xy=\dfrac34$,
$$x^4+y^4+3xy-2=(x^2+y^2)^2-2x^2y^2+3xy-2=(x^2+y^2)^2-\frac98+\frac94-2=0$$
and
$$x^2+y^2=\sqrt{\frac78}.$$ 
Anyway, we would have $(x-y)^2=x^2+y^2-2xy<0$, which is not possible.
Final answers: the solution points are $\pm\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)$ and $\pm(\sqrt2,-\sqrt2).$
