11. Find and classify the critical points of f(x,y) = x⁴ + y⁴ -4xy

f_x = 4x³ -4y = 0 Þ y = x³

f_y = 4y³ -4x = 0 Þ x = y³

Consequently x⁹ = x Þ x(x⁸ - 1) = 0 Þ x = 0, x = 1, or x = -1.

Going on to the second derivative test:

f_xx = 12x²

f_xy = -4

f_yy = 12y²

So

D = 144x²y² -16

Now, since y = x³, if x = 0 then y = 0 and D < 0, so the function has a saddle at (0,0).

If x = ± 1 then y = ± 1, D = 144-16 >0, and A = 12 so the function has local minimas at (± 1, ± 1)