Critical points of $g(x, y) = x^3+3xy^2-12x-6y$ 
Find the critical points of $g(x, y) = x^3+3xy^2-12x-6y$.

$\nabla g(x, y) = (3x^2+3y^2-12, 6xy-6)$ so $\nabla g(x, y) = 0 \implies  (3x^2+3y^2-12, 6xy-6) = (0,0)$
So $x^2+y^2 -4 = 0$ and $xy = 1$; combining and writing it in two different ways we get: 
$\begin{aligned} &\begin{cases} x^2+2xy+y^2 = 6xy \implies (x+y)^2 = 6 \implies x+y = \pm  \sqrt{2}\sqrt{3} \\ x^2-2xy+y^2 = 2xy \implies (x-y)^2 = 2 \implies x-y = \pm \sqrt{2} \end{cases}  \\& \implies (x,y) = (\pm \dfrac{1}{\sqrt{2}}(1+\sqrt{3}), \pm \frac{1}{\sqrt{2}}(\sqrt{3}-1))   ~ \text{are the critical points}.\end{aligned}$
This disagrees with the answer that wolfram alpha is giving me. What did I do wrong?

Your method can't find two other critical points. You successfully found that $x+y=\pm \sqrt{6}$ and $x-y=\pm \sqrt{2}$, but plus and minus signs don't need to coincide. Then four possibilities occur:


*

*$x=\dfrac{\sqrt{6}+\sqrt{2}}{2}$

*$x=\dfrac{\sqrt{6}-\sqrt{2}}{2}$

*$x=\dfrac{-\sqrt{6}+\sqrt{2}}{2}$

*$x=\dfrac{-\sqrt{6}-\sqrt{2}}{2}$


Here 2 and 3 are missing in your attempt.
