For f(x, y), find all values of x and y such that fx(x, y) = 0 and fy (x, y) = 0
simultaneously. f(x, y) = 15x^3-3xy+15y^3.

my calculations:
fx(x, y)=

fy(x, y)=

i got stuck there

(0,0)

(1/15,1/15)

Please let me know if you are expected to find complex solutions rather than just the real solutions I have provided.

Thanks kindly.

Step-by-step explanation:

Your derivatives look good.

Now we have to solve this system:

45x^2-3y=0

-3x+45y^2=0

Solving the first equation for y in terms of x:

45x^2-3y=0

Add 3y on both sides:

45x^2=3y

Divide both sides by 3:

15x^2=y

We are going to plug this into second equation:

-3x+45y^2=0 with y=15x^2

-3x+45(15x^2)^2=0

-3x+45×225x^4=0

Divide both sides by -3:

x-15×225x^4=0

x-3375x^4=0

Factor left hand side:

x(1-3375x^3)=0

This implies that x=0 or 1-3375x^3=0.

We can factor left hand side of second equation. It is a difference of cubes.

(1-15x)(1+15x+225x^2)=0

This means x=1/15 or 1+15x+225x^2=0.

Let's solve that second equation just a line above this.

1+15x+225x^2=0

a=225

b=15

c=1

First, I will find the discriminant to find out if the solution is even real.

b^2-4ac

(15)^2-4(225)(1)

225-4(225)

225(1-4)

225(-3)

A positive times a negative is negative so the discriminant tells us even if we find the solution, it is not real.

So we had x=0, 1/15 that we need to look at. I'm going to use the earlier equation I solved for y in terms of x to find the corresponding values per each x-solution.

15x^2=y with x=0:

15(0)^2=y

15(0)=y

0=y

So we have point (0,0) is a solution to the system.

15x^2=y with x=1/15:

15(1/15)^2=y

15/15^2=y

1/15=y

So we also have the point (1/15,1/15) is a solution to the system.