Answer
$y'=\dfrac{6y-x^2}{y^2-6x}$
Work Step by Step
Apply rule of differentiation: $y'=a'(x)+b'(x)$
Here, $a'(x)=m'(x)n(x)+m(x)n'(x)$
Given: $x^3+y^3=\frac{18xy}{d/dx}$
$3x^2+3y^2y'=18(x'y+xy')$
$3x^2+3y^2y'=18(y+xy')$
$3x^2+3y^2y'=18y+18xy'$
$-18xy'+3y^2y'=18y-3x^2$
$y'=\dfrac{18y-3x^2}{3y^2-18x}$
Hence, $y'=\dfrac{6y-x^2}{y^2-6x}$