A cleaner way to see this is with a pure symmetry argument, without directly computing any integral for $E[X^3]$.
Since $X$ is $Uniform(-1,1)$, its distribution is perfectly symmetric around 0: the random variable $-X$ has exactly the same distribution as $X$. Now look at how $Y$ is generated: given $X = x$, $Y$ is drawn uniformly from an interval centered at $x^2$. Since $(-x)^2 = x^2$, the conditional distribution of $Y$ given $X = x$ is identical to the conditional distribution of $Y$ given $X = -x$. In other words, $Y$'s distribution only depends on $x$ through $x^2$, so $Y$ is completely blind to the sign of $X$.
Now consider the pair $(X, Y)$ versus the pair $(-X, Y)$. Because of the symmetry just described, these two pairs have exactly the same joint distribution: for every value of $X$, the conditional law of $Y$ is unchanged if we flip the sign of $X$, and $X$ itself is equally likely to be positive or negative. This means:
$$E[XY] = E[(-X)Y] = -E[XY]$$The only number that equals its own negative is 0, so $E[XY] = 0$ follows immediately without any integration.
Also, $E[X] = 0$ since $Uniform(-1,1)$ is symmetric about the origin. So the covariance is:
$$Cov(X,Y) = E[XY] - E[X]E[Y] = 0 - 0 \cdot E[Y] = 0$$Both $X$ and $Y$ have finite positive variance ($X$ is a proper uniform variable, and $Y$'s conditional spread is a fixed constant plus the extra variation coming from $X^2$, so $\sigma_Y$ is finite and nonzero). With the numerator of the correlation formula equal to 0, the correlation itself must be 0:
$$correlation(X,Y) = \frac{Cov(X,Y)}{\sigma_X\sigma_Y} = \frac{0}{\sigma_X\sigma_Y} = 0$$ $$\boxed{0}$$