# What are the critical points of f(x,y) =x^3 + xy - y^3?

Apr 8, 2016

They are $\left(0 , 0\right)$ and $\left(\frac{1}{3} , - \frac{1}{3}\right)$

#### Explanation:

For $f \left(x , y\right) = {x}^{3} + x y - {y}^{3}$, we have

${f}_{x} = 3 {x}^{2} + y$
${f}_{y} = x - 3 {y}^{2}$.

We need to solve the system

$3 {x}^{2} + y = 0$
$x - 3 {y}^{2} = 0$.

The first equation gives us $y = - 3 {x}^{2}$.

Substituting for $y$ in the second equation gets us

$x - 3 {\left(- 3 {x}^{2}\right)}^{2} = 0$

$x - 27 {x}^{4} = 0$

$x \left(1 - 27 {x}^{3}\right) = 0$

$x = 0$ $\text{ }$ OR $\text{ }$ $x = \frac{1}{3}$.

Using $y = - 3 {x}^{2}$ from above we get critical points $\left(0 , 0\right)$ and $\left(\frac{1}{3} , - \frac{1}{3}\right)$.