finding critical points?

Question

how do i find the critical points of the function f(x,y) = x^3 - y^3 - 3xy + 4.

so far i managed to partially differentiate the function with respect to x and y to get
d/dx = 3x^2 - 3y

d/dy= -3y^2 - 3x

how do i go from here??

thx..

once i differenciate it implicitly, what do i do??

Answer 1

You have to use partial derivatives and put them to zero
fx=3x^2-3y=0
fy=-3y^2-3x=0
y=x^2 so -x^4-3x=0 so -x(x^3+3)=0
Which gives x=0 and x=-3^1/3
For x=0 y=0 so(0,0 is a critical point
For x=(-3)^1/3 y= 3^2/3 the other critical point
What happens at those points?
fxx=6x
fyy=-6y
fxy=-3
At(0,0) fxx*fyy-fxy^2 =-9 so(0,0) is a saddle point
at(-3^1/3,3^2/3)
fxx*fyy-fxy ^2=-3-9=-12<0 so it is also a saddle point

Answer 2

don't differentiate them separately, use implicit differentiation.