$\frac{d}{dx} \frac{x^2}{y}$
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)?