Answer
$y"=\frac{-14x}{y^5}$
Work Step by Step
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:
$y"=\frac{-14x}{y^5}$
