What is the minimum value of $x^2+y^2$ subject to $x^3+y^3+xy=1$? 
What is the minimum value of $x^2+y^2$ under the constraint $x^3+y^3+xy=1$? Please do not use partial differentials (multivariable calculus) or Lagrange multipliers. You can use elementary algebra or single variable calculus.

I plotted the graph of $x^3+y^3+xy=1$. It seems the minima occurs when $x=y$ but I don't know why that will be true. Any ideas?

Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative. 
Hence, $x^3+y^3+xy=1$ gives $2u(4u^2-3v^2)+v^2=1$ or $v^2=\frac{8u^3-1}{6u-1}$.
But $u^2\geq v^2$. 
Hence, $u^2\geq\frac{8u^3-1}{6u-1}$ or $\frac{2u^3+u^2-1}{6u-1}\leq0,$ which gives $\frac{1}{6}<u\leq u_1$,
where $u_1$ is a real root of the equation $2u^3+u^2-1=0$.
Hence, $x^2+y^2=4u^2-2v^2=4u^2-\frac{2(8u^3-1)}{6u-1}=\frac{2(4u^3-2u^2+1)}{6u-1}$, which is decreasing on $\left(\frac{1}{6},u_1\right]$,
which says that the answer is $\frac{2(4u_1^3-2u_1^2+1)}{6u_1-1}.$
We can get $u_1$ by the Cardano's formula.
