<s>[INST] Finding $\frac{d}{dx} \frac{x^2}{y}$ $$\frac{d}{dx} \frac{x^2}{y}$$
According to Wolframalpha
I "factor out constants"
$$\frac{\frac{d}{dx} x^2}{y}$$
Then I will get $\frac{2x}{y}$. Is that right? But $y$ is not a constant? What I did actually (quotient rule got me stuck)
The actual question is "Find $\frac{d^2y}{dx^2}$ of $2x^3 - 3y^2 = 8$"
I got 
$$\frac{dy}{dx} = \frac{x^2}{y}$$
Then 
$$\frac{d^2y}{dx^2} = \frac{y \cdot 2x - x^2 \cdot \frac{dy}{dx}}{y^2}$$
$$ = \frac{2xy - x^2 \cdot \frac{x^2}{y}}{y^2}$$
$$ = \frac{2xy^2 - x^4}{y^3}$$
Is this correct? It doesn't look like a "simple" answer (or whats in wolfram)?
 [/INST] Your first attempt (second in order presented) is correct. There is some simplification that you can do though. Since you will only ever be evaluating this expression for $(x,y)$ on the originally defined curve, you can use that relation to simplify. Some  "simplifications" would be 
$$
\begin{align*}
\frac{d^2y}{dx^2} & = \frac{2xy^2-x^4}{y^3}&\frac{d^2y}{dx^2} & = \frac{2xy^2-x^4}{y^3}\\
& = \frac{2x(3y^2)-3x^4}{3y^2y}&& = \frac{x(4y^2-2x^3)}{2y^3}\\
& = \frac{2x(2x^3-8)-3x^4}{(2x^3-8)y}&& = \frac{x(4y^2-(8+3y^2))}{2y^3}\\
& = \frac{x^4-16x}{(2x^3-8)y}&& = \frac{x(y^2-8)}{2y^3}\\\\
& = \frac{x(x^3-16)}{2(x^3-4)y}\\
\end{align*}
$$
This is only "simpler" in that lower powers of $y$ (respectively $x$) are used. 
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