Variance of product of dependent variables What is the formula for variance of product of dependent variables? 
In the case of independent variables the formula is simple:
$$ {\rm var}(XY) = E(X^{2}Y^{2}) - E(XY)^{2} = {\rm var}(X){\rm var}(Y) + {\rm var}(X)E(Y)^2 + {\rm var}(Y)E(X)^2 $$
But what is the formula for correlated variables?
By the way, how can I find the correlation based on the statistical data?
 A: Well, using the familiar identity you pointed out, 
$$ {\rm var}(XY) = E(X^{2}Y^{2}) - E(XY)^{2} $$ 
Using the analogous formula for covariance, 
$$ E(X^{2}Y^{2}) = {\rm cov}(X^{2}, Y^{2}) + E(X^2)E(Y^2) $$ 
and 
$$ E(XY)^{2} = [ {\rm cov}(X,Y) + E(X)E(Y) ]^{2} $$ 
which implies that, in general, ${\rm var}(XY)$ can be written as 
$$ {\rm cov}(X^{2}, Y^{2}) + [{\rm var}(X) + E(X)^2] \cdot[{\rm var}(Y) + E(Y)^2] - [ {\rm cov}(X,Y) + E(X)E(Y) ]^{2}  $$
Note that in the independence case, ${\rm cov}(X^2,Y^2) = {\rm cov}(X,Y) = 0$ and this reduces to 
$$ [{\rm var}(X) + E(X)^2] \cdot[{\rm var}(Y) + E(Y)^2] - [ E(X)E(Y) ]^{2}  $$
and the two $[ E(X)E(Y) ]^{2}$ terms cancel out and you get 
$$ {\rm var}(X){\rm var}(Y) + {\rm var}(X)E(Y)^{2} + {\rm var}(Y)E(X)^{2} $$ 
as you pointed out above. 
Edit: If all you observe is $XY$ and not $X$ and $Y$ separately, then I don't think there is a way for you to estimate ${\rm cov}(X,Y)$ or ${\rm cov}(X^2,Y^2)$ except in special cases (for example, if $X,Y$ have means that are known a priori)
A: This is an addendum to @Macro's very nice answer which lays out
exactly what needs to known in order to determine the variance of 
the product of two correlated random variables. Since 
\begin{align}
\operatorname{var}(XY) &= E\left[(XY)^2\right] - \left(E[XY]\right)^2
\tag{1}\\
&= E[(XY)^2] - \left(\operatorname{cov}(X,Y)+E[X]E[Y]\right)^2\\
&= E[X^2Y^2] - \left(\operatorname{cov}(X,Y)+E[X]E[Y]\right)^2\tag{2}\\
&= \left(\operatorname{cov}(X^2,Y^2)+E[X^2]E[Y^2]\right)
- \left(\operatorname{cov}(X,Y)+E[X]E[Y]\right)^2\tag{3}\\
\end{align}
where $\operatorname{cov}(X,Y)$, $E[X]$, $E[Y]$, $E[X^2]$, and
$E[Y^2]$  can be assumed to
be known quantities, we need to be able to determine the value of 
$E\left[X^2Y^2\right]$ in $(2)$ or $\operatorname{cov}(X^2,Y^2)$ in $(3)$.
This is not easy to do in general, but, as pointed out already, if
$X$ and $Y$ are independent random variables, then
$\operatorname{cov}(X,Y) = \operatorname{cov}(X^2,Y^2) = 0$.
In fact, dependence, not correlation (or lack thereof) is the
key issue. That we know that $\operatorname{cov}(X,Y)$ equals $0$
instead of some nonzero value does not, by itself, help in the 
least in our efforts are determining the value of 
$E\left[X^2Y^2\right]$ or $\operatorname{cov}(X^2,Y^2)$ even though it
does simplify the right sides of $(2)$ and $(3)$ a little.
When $X$ and $Y$ are dependent
random variables, then in at least one (fairly common
or fairly important) special
case, it is possible to find
the value of $E\left[X^2Y^2\right]$ relatively easily.
Suppose that $X$ and $Y$ are jointly normal random variables
with correlation coefficient $\rho$. Then, conditioned
on $X = x$, the conditional density of $Y$ is a normal
density with mean 
$E[Y] + \rho\left.\left.\sqrt{\frac{\operatorname{var}(Y)}{\operatorname{var}(X)}}
\right(x-E[X]\right)$ and variance $\operatorname{var}(Y)(1-\rho^2)$. Thus,
\begin{align}E[X^2Y^2 \mid X] &= X^2E[Y^2 \mid X]\\
&= X^2\left[\operatorname{var}(Y)(1-\rho^2)
+ \left(E[Y] + \rho\left.\left.\sqrt{\frac{\operatorname{var}(Y)}{\operatorname{var}(X)}}
\right(X-E[X]\right)\right)^2\right]
\end{align}
which is a quartic function of $X$, say $g(X)$, and the Law of Iterated
Expectation tells us that
$$E[X^2Y^2] = E\left[E[X^2Y^2\mid X]\right] = E[g(X)]\tag{4}$$
where the right side of $(4)$ can be computed from knowledge of the
3rd and 4th moments of $X$ -- standard results that can be found
in many texts and reference books
(meaning that I am too lazy to look them up
and include them in this answer).

Further addendum: In a now-deleted answer, @Hydrologist gives the variance of $XY$ as
$$\mathrm{Var}\left[xy\right] = \left(\mathrm{E}\left[x\right]\right)^2\mathrm{Var}\left[y\right] + \left(\mathrm{E}\left[y\right]\right)^2\mathrm{Var}\left[x\right] + 2\mathrm{E}\left[x\right]\mathrm{Cov}\left[x,y^2\right] + 2\mathrm{E}\left[y\right]\mathrm{Cov}\left[x^2,y\right]\\ + 2\mathrm{E}\left[x\right]\mathrm{E}\left[y\right]\mathrm{Cov}\left[x,y\right] +\mathrm{Cov}\left[x^2,y^2\right] - \left(\mathrm{Cov}\left[x,y\right]\right)^2 \tag{5}$$
and claims that this formula is from two papers published a half-century ago in JASA. This formula is an incorrect transcription of the results in the paper(s) cited by Hydrologist. Specifically, $\mathrm{Cov}\left[x^2,y^2\right]$ is a mistranscription of 
$E[(x-E[x])^2(y-E[y])^2]$ in the journal article, and similarly for $\mathrm{Cov}\left[x^2,y\right]$ and $\mathrm{Cov}\left[x,y^2\right]$.
