Answer
$y' = \dfrac{3y^2-3x^2}{3y^2-6xy}$
Work Step by Step
In order to derivate this function you have to apply implicit differentation method.
First, take the function to it's $f(x) = 0$ form
$x^3+y^3-3xy= 0$
Then derivate the whole equation. Rember the put y' every time you derivate y
$3x^2+3y^2y' - (3y^2+6xyy')= 0$
*Note: Here you have to apply the product rule
Solve for y' and you have the answer
$y'(3y^2-6xy) = 3y^2-3x^2$
$y' = \dfrac{3y^2-3x^2}{3y^2-6xy}$
