can somebody please check the following for me. many thanks

Find an Expression for DY/DX (I use D as i will use d for partial derivatives)

When: x^3 + y^3 - 2x^2 y = 0

i) Assign;

z = x^3 + y^3 - 2x^2 y = 0 (eqn 1)

we also have Ez = dz/dx . Ex + dz/dy . Ey ( Where E = Delta/Change)

ii) Divide Both Sides By Ex

which then equates to ;

Ez/Ex = dz/dx + dz/dy . Ey / Ex (eqn 2)

and when Ex approaches zero eqn1 becomes;

DZ/DX = dz/dx + dz/dy . DY/DX (eqn 3)

iii) we can now find the partial derivatives;

dz/dx = 3x^2 - 4xy

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

and since z = zero we can now substitue these back into eqn 3 to get;

0 = 3x^2 - 4xy + (3y^2 - 2x^2) . DY/DX

- 3x^2 + 4xy = (3y^2 - 2x^2) . DY/DX

DY/DX = (-3x^2 + 4xy) / (3y^2 - 2x^2)

Think thats not far off, sorry for changing the script and making it confursing.

Thanks again