## Calculus: Early Transcendentals 8th Edition

$y"=\frac{-14x}{y^5}$
Find y' by taking derivative of both sides of the equation: $3x^2-3y^2\times dy/dx=0$ $y'=\frac{x^2}{y^2}$ Find y": $y"=\frac{y^2(2x)-(x^2)(2y\times y')}{y^4}$ Plug y' into y": $y"=\frac{2xy^2-2yx^2\times \frac{x^2}{y^2}}{y^4}$ Simplify: $y"=\frac{2xy^4-2yx^4}{y^6}=\frac{2xy^3-2x^4}{y^5}$ Substitute $y^3=x^3-7$ and simplify: $y"=\frac{-14x}{y^5}$