Let $ g\left(x,y\right) = x^3 -9xy^2 +9y^4 + 23$
and the gradient of that function:
$$ \nabla g\left(x,y\right) = \left( 3x^2 -9y^2 \space , \space -18xy + 36y^3 \right)^T$$
To solve for $$ \nabla g\left(x,y\right) = \left(0,0\right)$$ I did the following steps:
$$ 3x^2 - 9y^2 = 0 \\ -18xy + 36y^3 = 0$$
$$ x^2 - 3y^2 = 0 \\ -xy + 2y^3 = 0$$
$$ x^2 = 3y^2 \rightarrow (x = 0) \land (y = 0)$$ Since anything squared is positive and setting those values to zero is the only way to satisfy this equation
$$ 2y^3 = xy $$ This is where I'm kind of stuck. If $ x= 0 $ then $ y =0 $. But when I try to verify my answer with a online tool I see that I missed a critical point.
The critical point which I found at $(0,0)^T$ seems to be a saddle point since the Hesse Matrix $H g(x,y)$ is indefinite at that point.
What mistake did I make ?
If $y \ne 0$, then we get from $2y^3=xy$ that $x=2y^2$. From $x^2=3y^2$ we then derive
$4y^4=3y^2$.
Therefore $4y^2=3$.
Can you proceed ?