Answer
$\frac{d^{2}y}{dx^{2}} = -\frac{21}{16y^{3}}$
Work Step by Step
$3x^{2}-4y^{2} = 7$
Differentiating implicitly :
$6x - 8y\frac{dy}{dx} = 0$
$6x = 8y\frac{dy}{dx}$
$\frac{6x}{8y} = \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{3x}{4y}$
Again differentiating implicitly :
$\frac{d^{2}y}{dx^{2}} = \frac{3}{4}[\frac{y(1)-x(\frac{dy}{dx})}{y^{2}}]$
$\frac{d^{2}y}{dx^{2}} = \frac{3}{4y^{2}}(y-x\frac{dy}{dx})$
As $\frac{dy}{dx} = \frac{3x}{4y}$
So,
$\frac{d^{2}y}{dx^{2}} = \frac{3}{4y^{2}}[y-x(\frac{3x}{4y})]$
$\frac{d^{2}y}{dx^{2}} = \frac{3}{4y^{2}}(y-\frac{3x^{2}}{4y})$
$\frac{d^{2}y}{dx^{2}} = \frac{3}{4y^{2}}(\frac{4y^{2}-3x^{2}}{4y})$
$\frac{d^{2}y}{dx^{2}} = \frac{3}{16y^{3}}[-(3x^{2}-4y^{2})]$
As from the question statement we know that $3x^{2}-4y^{2} = 7$
So,
$\frac{d^{2}y}{dx^{2}} = \frac{3}{16y^{3}}(-7)$
$\frac{d^{2}y}{dx^{2}} = -\frac{21}{16y^{3}}$
