Is $\frac{d}{d^2x}$ a valid differential operator? Is $\frac{d}{d^2x}$ a valid differential operator? If it is would it represent something along the lines of the change in y as the change in x changes? How would it be evaluated?
 A: $$\frac{dy}{d^2x}=\frac{dy}{dx}\cdot\frac{1}{dx}$$
But $\frac{1}{dx}$ has no meaning, thusly neither does $\frac{dy}{d^2x}$
On the other hand, we have $$\frac{dy}{d(x^2)}=\frac{dy}{dx}\cdot\frac {1}{2x}$$ 
(can you show why?)
A: First question - is it a valid operator?  Yes indeed.  There's nothing incorrect about it.  See the paper "Extending the Algebraic Manipulability of Differentials" for more justification of using differentials in this way.
The way to evaluate it depends on what you consider a solution.  You are basically taking the differential and dividing by $d^2x$.  So, here is a way of looking at this:
$$y = x^3 \\
dy = 3x^2\,dx \\
\frac{dy}{d^2x} = \frac{3x^2\,dx}{d^2x}$$
So, that is a valid formulation, but it doesn't actually tell you much.  Alternatively, if we solved for $x$, we could get:
$$
x = y^{\frac{1}{3}} \\
dx = \frac{1}{3}y^{-\frac{2}{3}}\,dy \\
d^2x = -\frac{2}{9}y^{-\frac{5}{3}}\,dy^2 + \frac{1}{3}y^{-\frac{2}{3}}d^2y 
$$
Therefore, $\frac{dy}{d^2x}$ would be
$$
\frac{dy}{d^2x} = \frac{3x^2\,dx}{-\frac{2}{9}y^{-\frac{5}{3}}\,dy^2 + \frac{1}{3}y^{-\frac{2}{3}}d^2y}
$$
Again, I'm not sure what that actually tells you, but there's nothing invalid about it.  However, note that when using differentials algebraically, $\frac{d^2y}{dx^2}$ is not the second derivative.  Instead it is $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$.  So, the notation is a little different than you may be used to for higher-order differentials, but they do indeed work algebraically!
