(2) find the critical points of $f(x,y)=x^{2}-x^{4}+xy + y^{2}$.

Explanation:

Step1: Calculate partial - derivatives

First, find the first - order partial derivatives of (f(x,y)). The partial derivative with respect to (x) is: [f_x=\frac{\partial f}{\partial x}=2x - 4x^{3}+y] The partial derivative with respect to (y) is: [f_y=\frac{\partial f}{\partial y}=x + 2y]

Step2: Set partial - derivatives equal to zero

Set (f_x = 0) and (f_y=0). From (f_y=x + 2y=0), we get (x=-2y). Substitute (x = - 2y) into (f_x): [2(-2y)-4(-2y)^{3}+y = 0] [ - 4y+32y^{3}+y = 0] [32y^{3}-3y = 0] Factor out (y): (y(32y^{2}-3)=0)

Step3: Solve for (y)

Case 1: If (y = 0), then from (x=-2y), (x = 0). Case 2: If (32y^{2}-3=0), then (y^{2}=\frac{3}{32}), so (y=\pm\frac{\sqrt{3}}{4\sqrt{2}}=\pm\frac{\sqrt{6}}{8}) When (y=\frac{\sqrt{6}}{8}), (x=-2y=-\frac{\sqrt{6}}{4}) When (y =-\frac{\sqrt{6}}{8}), (x=-2y=\frac{\sqrt{6}}{4})

Answer:

The critical points are ((0,0)), ((-\frac{\sqrt{6}}{4},\frac{\sqrt{6}}{8})), and ((\frac{\sqrt{6}}{4},-\frac{\sqrt{6}}{8}))