## University Calculus: Early Transcendentals (3rd Edition)

$y'=\dfrac{6y-x^2}{y^2-6x}$
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, $y'=\dfrac{6y-x^2}{y^2-6x}$