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**Soroban** Hello, slevvio!

I don't get their answer either . . .

$\displaystyle \text{Differentiate implicitly: }\;2x + 8y\frac{dy}{dx}\:=\:0\quad\Rightarrow\quad\frac{dy }{dx}\:=\:-\frac{x}{4y}\:=\:-\frac{1}{4}\cdot\frac{x}{y}$

$\displaystyle \text{Differentiate: }\;\frac{d^2y}{dx^2} \;=\;-\frac{1}{4}\left[\frac{y\!\cdot\!1 - x\!\cdot\!\frac{dy}{dx}}{u^2}\right] \;=\;-\frac{1}{4}\cdot\frac{y - x\frac{dy}{dx}}{y^2}$

$\displaystyle \text{Since }\frac{dy}{dx} \:=\:-\frac{x}{4y}\text{, substitute: }\;\frac{d^2y}{dx^2} \;=\;-\frac{1}{4}\cdot\frac{y - x\left(-\frac{x}{4y}\right)}{y^2}$

$\displaystyle \text{Multiply top and bottom by }4y\!:\;\;\frac{d^2y}{dx^2}\;=\;-\frac{1}{4}\cdot\frac{4y^2 + x^2}{4y^3} $

$\displaystyle \text{So we have: }\;\frac{d^2y}{dx^2} \;=\;- \frac{{\color{blue}x^2+4y^2}}{16y^3}$

$\displaystyle \text{According to the original equation: }\:x^2+4y^2 \:=\:{\color{blue}9}$

$\displaystyle \text{Therefore: }\:\frac{d^2y}{dx^2} \:=\:-\frac{9}{16y^3}$