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)?

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. 
