Min, max and saddle points of a function I want to find all local minima, maxima and saddle points of the function $f(x,y)=(x-y)(1-xy)$
Therefore I wanted to set the partial derivatives to zero to get all possible points.
First I calculated the derivatives $f_x(x,y) = y^2-2xy+1$ and $f_y(x, y) = -x^2+2xy-1$
From $f_x$ I got $x=\frac{y^2+1}{2y}$ and I used it in $f_y$ to be able to determine the values of $y$.
$-\frac{y^2+1}{2y}^2+2y\frac{y^2+1}{2y}-1=0$
Then I got $y^4-\frac{10}{3}y^2-\frac{1}{3}=0$
I substituted $z = y^2$ and got $z^2-\frac{10}{3}z-\frac{1}{3}=0$
When I wanted to solve this, it leads to $z_1= \frac{5+\sqrt(28)}{3}$ and $z_2= \frac{5-\sqrt(28)}{3}$. As I am not allowed to use a calculator, I think that I've done something wrong as it's for example not very easy for me to determine the result of $\sqrt(28)$ without using one, but I'm not able to find what I may have done wrong.

I finally found the error in a sign. When correcting it, it leads to
$y^4-\frac{2}{3}y^2-\frac{1}{3}=0$
which leads to the values  $y_1 = 1$ and $y_2 = -1$.
Then the possible points are $(1,1)$ and $(-1,-1)$.
The point $(1,1)$ is a minimum, $(-1,-1)$ is a maximum and there are no saddle points.
