Well I tried out this equation(x^3-2x^2y+3xy^2 = 38, and i got dy/dx = 4xy+3y^2/-2x^2+6xy. I know it is incorrect. Can someone walk me through this so I can correct my mistake? I'm studying for a test on monday
$\displaystyle \frac{d}{dx}\left[x^3 - 2x^2y + 3xy^2 = 38\right]
$
$\displaystyle 3x^2 - 2x^2 \frac{dy}{dx} - 4xy + 6xy \frac{dy}{dx} + 3y^2 = 0$
$\displaystyle \frac{dy}{dx}\left(6xy - 2x^2\right) = 4xy - 3x^2 - 3y^2$
$\displaystyle \frac{dy}{dx} = \frac{4xy - 3x^2 - 3y^2}{6xy - 2x^2}$