# How do you find critical points of multivariable function f(x,y) =x^3 + xy - y^3?

Apr 11, 2015

Several notations and explanations are available. Here's one:

Find the partial derivatives, set them equal to zero and solve the resulting system of equations.

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

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

From the first equation: $y = - 3 {x}^{2}$.

So, we need: $x - 3 {\left(- 3 {x}^{2}\right)}^{2} = 0$

$x - 27 {x}^{4} = 0$ when $x = 0$ , in which case $y = - 3 {\left(0\right)}^{2} = 0$

and also when $x = \frac{1}{3}$ in which case $y = - \frac{1}{3}$

The critical points are: $\left(0 , 0\right)$ and $\left(\frac{1}{3} , - \frac{1}{3}\right)$.

(I've heard that there is an alternative terminology that would find the values of $f$ and say that critical points are points in 3-space: $\left(0 , 0 , 0\right)$ and $\left(\frac{1}{3} , - \frac{1}{3} , - \frac{1}{27}\right)$)