Can you please show me how to find the derivative of the given function: (2x-3y)^3 = x^2 - y?

(2x-3y)^3 = x^2-y

Take deribvative with respect to x
3(2x-3y)^2 ( 2 - 3 dy/dx) = 2x - dy/dx
6(2x-3y)^2 -9(2x-3y)^2 dy/dx = 2x - dy/dx

dy/dx (1-9(2x-3y)^2) = 2x - 6(2x-3y)^2
dy/dx = ( 2x - 6(2x-3y)^2 ) / ( 1 - 9(2x-3y)^2)
dy/dx = ( 2x - 6(4x^2-12xy+9y^2) ) /( 1-9(4x^2-12xy+9y^2))

dy/dx = (2x-24x^2+72xy-54y^2) /(1-36x^2+108xy-81y^2)

(2x - 3y)³ = x² - y
8x³ - 36x²y + 54xy² - 27y³ = x² - y
(d/dx)(8x³ - 36x²y + 54xy² - 27y³) = (d/dx)(x² - y)
24x² - 36x²(dy/dx) - 72xy + 108xy(dy/dx) + 54y² - 81y² = 2x - (dy/dx)
24x² - 36x²(dy/dx) - 72xy + 108xy(dy/dx) - 27y² = 2x - (dy/dx)
36x²(dy/dx) - 108xy(dy/dx) - (dy/dx) = 24x² - 72xy - 27y² + 2x
(36x² - 108xy - 1)(dy/dx) = 24x² - 72xy - 27y² + 2x
dy/dx = (24x² - 72xy - 27y² + 2x)/(36x² - 108xy - 1)

with respect to x or y?

Assuming you want dy/dx
y' = dy/dx
3(2-3y')(2x-3y)^2 = 2x - y'
Now all you have to do is solve for y'

F(x,y)=(2x-3y)³-x²+y=0

We can use Dini theorem (implicit function theorem) to find derivative.

dy/dx=-∂F(x,y)/∂x/∂F(x,y)/∂y=

=(-12y-8x+2x)/(18y-12x+1)=

=(12y-6x)/(18y-12x+1)

Check. http://www.wolframalpha.com/input/?i=derivative+(2x-3y)%5E2%2By-x%5E2%3D0

OK!

3(2 - 3y')(2x - 3y)^2 = 2x - y'
y' = (24x^2 - 72y + 54y^2 - 2x) / (4x^2 - 12y + 9y^2 + 1)

"chain rule" .....3 ( 2x - 3y)² { 2 - 3 dy/dx} = 2x - dy/dx.....solve for dy/dx