Consider the equation y3−32x2=1.

Step 1
Consider the equation
${\displaystyle y^3-\frac{3}{2}{x}^2=1}$
Let us find the second derivative implicitly.
Step 2
Differentiate each term with respect to x
${\displaystyle \frac{d}{dx}(y^3-\frac{3}{2}{x}^2)=\frac{d}{dx}}$...(1)
${\displaystyle \frac{d}{dx}\left(y^3\right)-\frac{3}{2}\frac{d}{dx}\left(x^2\right)=0}$
${\displaystyle 3y^2\frac{dy}{dx}-3x=0}$
${\displaystyle \frac{dy}{dx}=\frac{x}{y^2}\to \left(1\right)}$
Step 3
Differentiate second time with respect to x
${\displaystyle \frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left(\frac{x}{y^2}\right)}$
${\displaystyle =\frac{\frac{d}{dx}\left(x\right)y^2-\frac{d}{dx}\left(y^{2}x\right)}{{\left(y^2\right)}^2}}$
${\displaystyle =\frac{1\cdot y^2-2y\frac{dy}{dx}x}{y^4}}$
${\displaystyle =\frac{y-2x\frac{dy}{dx}}{y^3}}$
Step 4
From equation (1) we have ${\displaystyle \frac{dy}{dx}=\frac{x}{y^2}}$. Substituting this we get
${\displaystyle \frac{d^{2}y}{{dx}^2}=\frac{y-2x\left(\frac{x}{y^2}\right)}{y^3}}$
${\displaystyle =\frac{y^3-2x^2}{y^5}}$

Answer is given below (on video)