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)?

$$
2x^3-3y^2 = 8.
$$
Differentiate both sides with respect to $x$:
$$
6x^2 - 6yy' = 0.
$$
Solve for $y'$:
$$
y' = \frac{x^2}{y}.
$$
Differentiate again with respect to $x$:
$$
y'' = \frac{y(2x)- x^2y'}{y^2}.
$$
Put $x^2/y$ in place of $y'$ and simplify:
$$
y'' = \frac{2xy - x^2\frac{x^2}{y}}{y^2} = \frac{2xy^2 - x^4}{y^3}.
$$
In a sense, you're done now, but notice that in place of $y^2$ you can put $(2x^3-8)/3$:
$$
y'' = \frac{2x(2x^3-8)/3 - x^4 }{y(2x^3-8)/3} = \frac{3x^4 - 16x}{y(2x^3-8)}.
$$
Maybe depending on the purpose, you might prefer to put $(8+3y^2)/2$ in place of $x^3$:
$$
y'' = \frac{2xy^2-x(8+3y^2)/2}{y^3} = \frac{4xy^2-8x-3xy^2}{2y^3}.
$$
