Find and classify all the stationary points of f(x,y)= 3(x^2)y - 3(x^2) - 3(y^2) + (y^3) + 2?

Question

Here's my work on this problem.
(df/dx)=6xy-6x = y-1....(1)
(df/dy)=3(x^2)-6y+(3y^2) = (x^2)-2y+(y^2)......(2)
since (df/dx)=0=(df/dy)
hence from (1) we get y=1, substituting this in (2), i get x=1.
Hence (1,1) is the stationary piont.
have i done this right till here?
Now this is the part where i'm going wrong,
A=(d^2f/dx^2)= 0, B=(d^2f/dxdy)=1 or 2x, C=(d^2f/dy^2)= 2y-2.
I need to find D = B^2 - AC, thus
if D>0, (1,1) is a saddle point,
if D<0, a maximum at (1,1) when A<0 and minimum at (1,1) when A>0.
if D=0, does not give any conclusion.

Answer 1

Y!A ate some of your formulas. I recommend putting spaces between your operations in order to prevent that.

df/dx should by (y-1)(6x), which is zero when y=1 or x=0.
df/dy should be 3x^2+3y^2 -6y. When y=1, this is 3x^2-3=3(x+1)(x-1), so x could be 1 or -1. Alternatively, when x=0, this is 3y^2-6y = 3y(y-2), so y could be 0 or 2.

So, I get the points (-1,1), (1,1), (0,0), (0,2).

Your second derivatives look a bit off. Without seeing the first derivatives you're working from, I can't tell you what the problem is. I'm going to use the notation f_xx for the second partial with respect to x, f_xy for the mixed second partial, and f_yy for the second partial with respect to y.

To find f_xx, differentiate (y-1)(6x) with respect to x. I get f_xx = 6(y-1).
f_xy = 6x
f_yy = 6(y-1)

At (-1,1), A=C=0 and B=-6, so D = B^2-4AC = 36 > 0; saddle point.
At (1,1), similar.
At (0,2), A=C=6 and B=0, so D<0; since A>0, this is a relative minimum.
At (0,0), A=C=-6 and B=0, so rel. max.