zero covariance but not independent - normally distributed random variable $X$ and $X^2$

This is one of my homework question, which the answer sheet has already been given out. However, I still don't understand it.

Exercise 1.1. It is well known that for two normal random variables, zero covariance implies independence. Why does this not apply to the following situation:

$X ∼ N(0, 1),\, \text{Cov}(X, X^2) = E[X^3] − E[X]\times E[X^2] = 0 − 0 = 0$

but obviously $X^2$ is totally dependent on X?

So, I assume that I have to show :

$f (x_1, x_2) = f_{X_1} (x_1) * f_{X_2} (x_2)$

for the random variables $X$ and $X^2$ to be not independent. The problem is that I do not know how to compute the joint density function $f (x_1, x_2)$ for the two random variables $X$ and $X^2$.

First of all, is this correct?

$f_{X_1} (x) =$ $1\over{\sqrt{2\pi\sigma^2}}$ $e^{- x^2 \over 2\sigma^2}$

$f_{X_2} (x) =$ [$1\over{\sqrt{2\pi\sigma^2}}$ $e^{- x^2 \over 2\sigma^2}]^2$.

Then, how do I compute the joint density function $f (x_1, x_2)$ ?

Since $X^2$ isn't a normal random variable, there's no contradiction.