Answer
$$\left( {0,0} \right),\,\,\,\left( {3, - 3} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{{{x^3}}}{3} - \frac{{{y^3}}}{3} + 3xy \cr
& {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{{x^3}}}{3} - \frac{{{y^3}}}{3} + 3xy} \right] \cr
& {f_x}\left( {x,y} \right) = {x^2} + 3y \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{{x^3}}}{3} - \frac{{{y^3}}}{3} + 3xy} \right] \cr
& {f_y}\left( {x,y} \right) = - {y^2} + 3x \cr
& \cr
& {\text{Set }}{f_x}\left( {x,y} \right) = 0{\text{ and }}{f_y}\left( {x,y} \right) = 0 \cr
& {x^2} + 3y = 0{\text{ and }} - {y^2} + 3x = 0 \cr
& {\text{From the equation }}{x^2} + 3y = 0 \cr
& y = - \frac{{{x^2}}}{3} \cr
& {\text{Substituting }} - \frac{{{x^2}}}{3}{\text{ for }}y{\text{ into }} - {y^2} + 3x = 0 \cr
& - {\left( { - \frac{{{x^2}}}{3}} \right)^2} + 3x = 0 \cr
& - {\frac{x}{9}^4} + 3x = 0 \cr
& {x^4} - 27x = 0 \cr
& x\left( {{x^3} - 27} \right) = 0 \cr
& {x_1} = 0{\text{ and }}{x_2} = 3 \cr
& {\text{We have that }}y = - \frac{{{x^2}}}{3}.{\text{ So}}{\text{, }} \cr
& {x_1} = 0 \Rightarrow {y_1} = 0 \cr
& {x_2} = 3 \Rightarrow {y_2} = - 3 \cr
& {\text{The critical points occur at}} \cr
& \left( {0,0} \right),\,\,\,\left( {3, - 3} \right) \cr} $$
