dy/dx implicit differentiation of x^3 - y^3 + 3(x^2)y = 17

[AhItsCalculus]

I have to find dy/dx of:

x^3 - y^3 + 3(x^2)y = 17

I keep getting different answers. The answer I get most is:

(-3x^2 - 6xy) / (-3y^2 + 3x^2) = dy/dx

Is this correct? If not, where did I go wrong?

process:

3x^2 - 3y^2(dy/dx) + 3(2xy + x^2(dy/dx)) = 0

-3y^2 (dy/dx) + 3x^2 (dy/dx) = -3x^2 - 6xy

(-3x^2 - 6xy) / (-3y^2 + 3x^2) = dy/dx

 

[skeeter]

\(\displaystyle \L x^3 - y^3 + 3x^2y = 17\)

\(\displaystyle \L 3x^2 - 3y^2\frac{dy}{dx} + 3x^2\frac{dy}{dx} + 6xy = 0\)

\(\displaystyle \L x^2 - y^2\frac{dy}{dx} + x^2\frac{dy}{dx} + 2xy = 0\)

\(\displaystyle \L x^2 + 2xy = y^2\frac{dy}{dx} - x^2\frac{dy}{dx}\)

\(\displaystyle \L x^2 + 2xy = \frac{dy}{dx}(y^2 - x^2)\)

\(\displaystyle \L \frac{x^2 + 2xy}{y^2 - x^2} = \frac{dy}{dx}\)

 

[AhItsCalculus]

Thank you sooo much!