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)?
 A: 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. 
A: $y$ is assumed to be a function of $x$ here, it's not a constant.  So your first solution is not correct (note WA interpreted the input as taking a partial derivative with respect to $x$, which is a different from what you want). 
You need to use the "implicit differentiation" method to find the derivatives  (you did this correctly in your second method):
$y$ is defined implicitly as a function of $x$ by the equation
$$
2x^3-3y^2=8.
$$
Let's find the first derivative. To find $dy\over dx$, differentiate both sides of the above with respect to $x$ keeping in mind that $y$ is a function of $x$:
$$
{d\over dx}(2x^3-3y^2)={d\over dx} 8.
$$
$$
6x^2-6y {dy\over dx}=0.
$$
Note that we needed to use the chain rule to find ${d\over dx} 3y^2$.
Solving for ${dy\over dx}$ gives $$ {dy\over dx } ={x^2\over y }.$$
To find  ${d^2y\over dx^2}$, differentiate both sides of the above  with respect to $x$:  
$$
{d\over dx}{dy\over dx } ={d\over dx}{x^2\over y }.
$$
$$
{d^2y\over dx^2}= {2x y-{dy\over dx} x^2\over y^2} ;
$$
simplifying the right hand of the above side leads to your solution.
Sometimes answers aren't simple; such is life...
A: The reason WolframAlpha treats $y$ as a constant here is that as far as it knows, $y$ is constant with respect to $x$, and $\frac{d}{dx}$ denotes taking the derivative with respect to $x$. However, in your problem $y$ is a function of $x$ rather than an independent variable, so your approach is the correct one.
A: $$
2x^3-3y^2 = 8.
$$
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 $x^3$:
$$
y'' = \frac{2xy^2-x(8+3y^2)/2}{y^3} = \frac{4xy^2-8x-3xy^2}{2y^3}.
$$
