How could I find the covariance for $X$ and $Y$ in this case? If $X \sim U(-1, 1)$ (so $X$ is uniformly distributed between $-1$ and $1$) and $Y = X^2$, what is the covariance between $X$ and $Y$? Are they independent?
So the formula for covariance is:
$\operatorname{Cov}(X,Y) = E(XY) - E(X)E(Y)$
Since $Y = X^2$
$E(XY) = E(X^3)$
and
$E(Y) = E(X^2)$
and
$\operatorname{Cov}(X, Y) = E(X^3) - E(X)E(X^2)$
is this the correct way to go about solving the problem?
 A: That is indeed correct.  It might also help to draw a simple picture of the curve on which the point $(X,Y)$ is constrained to lie, and see if that tells you something about what to expect the covariance to be.  The easier question is whether their independent: Notice that if you know what $X$ is, that tells you what $Y=X^2$ is.
A: Yes. That is what you need to do. Do you know what to do next? 
Hint:

 Obviously $\mathsf E(X)=0$, so $\mathsf E(X)\mathsf E(X^2)=0$.  So can you find $\mathsf E(X^3)$ ? (note: you can do it without calculation).

A: You're correct. $$E[X^3] = E[X] = 0$$ and hence $$\operatorname{Cov}(X,X^2) = 0$$
They are uncorrelated but not independent:
Observe that $$P(X \in (\frac{-1}{2},\frac{1}{2}), X^2 \in (\frac{-1}{2},\frac{1}{2})) \ne P(X \in (\frac{-1}{2},\frac{1}{2})) P(X^2 \in (\frac{-1}{2},\frac{1}{2}))$$
This is a classic example of a counterexample to the converse of independent implies uncorrelated. Uncorrelated means linearly independent: $$E[XY] = E[X] E[Y]$$
Independent means:
$$E[f(X)g(Y)] = E[f(X)] E[g(Y)]$$
where $f$ and $g$ are bounded and Borel-measurable. So we see that uncorrelated is what we get if we assume independence and use identity functions for $f$ and $g$. Independence is linear independence, quadratic independence, cubic independence, ..., trigonometric independence, exponential independence and so on.
