1. The problem statement, all variables and given/known data Find all stationary points of the function G(x, y) = (x^3)*e^(−x^2−y^2) 2. Relevant equations fx=0 and fy=0 3. The attempt at a solution Gx = 3x^2*e^(-x^2-y^2) +x^3(-2x)e^(-x^2-y^2) = e^(-x^2-y^2)(3x^2-2x^4) Gx = 0 implies 3x^2-2x^4=0 x^2(3-2x^2)=0 hence x =0 ,+or- (3/2)^(1/2) Gy = (-2y)(x^3)(e^(-x^2-y^2)) Gy=0 implies (-2y)(x^3)(e^(-x^2-y^2))=0 y=0 x = 0 (+or- (3/2)^(1/2), 0) are the two stationary points according to wolfram alpha but I don't understand why (o,a), where a is a real number, isn't a solution since Gx=Gy=0 when x = 0. Any help would be appreciated.