The function is
#f(x,y)=3x^2y+2y^3-3xy#
The partial derivatives are
#(delf)/(delx)=6xy-3y#
#(delf)/(dely)=3x^2+6y^2-3x#
Let #(delf)/(delx)=0# and #(delf)/(dely)=0#
Then,
#{(6xy-3y=0),(3x^2+6y^2-3x=0):}#
#<=>#, #{(3y(2x-1)=0),(x^2+2y^2-x=0):}#
#<=>#, #{(y=0),(x=1/2 ; 0 ; 1):}#
The points are #(0,0)#, #(1/2, 0)# and #(1,0)#
The second partial derivatives are
#(del^2f)/(delx^2)=6y#
#(del^2f)/(dely^2)=12y#
#(del^2f)/(delxdely)=6x-3#
#(del^2f)/(delydelx)=6x-3#
The Hessian matrix is
#Hf(x,y)=(((del^2f)/(delx^2),(del^2f)/(delxdely)),((del^2f)/(delydelx),(del^2f)/(dely^2)))#
The determinant of the matrix is
#D(x,y)=det(H(x,y))=|(6y,6x-3),(6x-3,12y)|#
#=6y*12y-(6x-3)^2#
#=72y^2-(6x-3)^2#
Therefore,
#D(0,0)=-9#
#D(1/2,0)=0#
#D(1,0)=-9#
For the point #(1/2,0)# the test is inconclusive.
There are saddle points at #(0,0)# and #(1,0)#.
