Q: Prove that $\text{Corr}(X^2,Y^2)=\rho^2$ where $X,Y$ are jointly $N(0,1)$ variables with correlation $\rho$ 
Consider jointly  normally distributed random variables $X,Y\sim N(0,1)$ that have $\text{Corr}(X,Y)=\rho$. 
  Show that $\text{Corr}(X^2,Y^2)=\rho^2$. 

(Hint: Consider $X,U\sim N(0,1)$ where they are independent then we have $Y=\rho X+\sqrt{1-\rho^2}U$). 
I sense that a transform is required for both $X$ and $Y$ but not sure where to go next.  
edit: I am more interested in knowing how the hint is used. 

A: Hint:
Let $(X,Y)$ be jointly normal variables with zero means and unit variances and $\text{Corr}(X,Y)=\rho$.
By definition, \begin{align}\text{Corr}(X^2,Y^2)=\frac{\text{Cov}(X^2,Y^2)}{\sqrt{\text{Var}(X^2)\text{Var}(Y^2)}}\end{align}
where $\text{Cov}(X^2,Y^2)=\mathbb E(X^2Y^2)-\mathbb E(X^2)\mathbb E(Y^2)$, and
$\text{Var}(X^2)=\mathbb E(X^4)-(\mathbb E(X^2))^2=\text{Var}(Y^2)$.
For finding $\mathbb E(X^2Y^2)$ quickly, note that $\mathbb E(X^2Y^2)=\mathbb E(\mathbb E(X^2Y^2\mid X))=\mathbb E(X^2\,\mathbb E(Y^2\mid X))$.
And we know that $Y\mid X\sim\mathcal{N}(\rho X,1-\rho^2)$.
So, $\mathbb E(Y^2\mid X)=\text{Var}(Y\mid X)+(\mathbb E(Y\mid X))^2=\cdots$.
I think you can find the moments now. 

A: You ask how to use the hint.  One way is to focus on computing the covariances that go into the correlation formula.  There are two.  I will do the algebra for you to reduce the problem to simpler calculations requiring a little statistical thought.
The easy covariance calculation (because it involves only one variable at a time and both variables have standard Normal distributions and we know their first four moments are $0,1,0,3$) is $$\operatorname{Var}(X^2) = \operatorname{Var}(Y^2) = E[(Y^2)^2]-E[Y^2]^2=3-1=2.\tag{*}$$
This implies you would like to prove
$$\operatorname{Cov}(X^2,Y^2) = \sqrt{\operatorname{Var}(X^2)}\sqrt{\operatorname{Var}(Y^2)} \operatorname{Cor}(X^2,Y^2) = 2\rho^2.\tag{**}$$
To compute this, let's just blindly apply the hint by making the substitution for $Y,$ expanding $Y^2$ algebraically, and exploiting the linearity of $\operatorname{Cov}$ in its second argument to break the resulting expression into three simpler ones:
$$\eqalign{
\operatorname{Cov}(X^2,Y^2) &= \operatorname{Cov}(X^2, (\rho X + \rho^\prime U)^2) \\
&= \operatorname{Cov}(X^2, \rho^2 X^2 + 2\rho\rho^\prime XU + (\rho^\prime)^2 U^2)\\
&= \rho^2 \operatorname{Cov}(X^2, X^2) + 2\rho\rho^\prime \operatorname{Cov}(X^2, XU) + (\rho^\prime)^2 \operatorname{Cov}(X^2, U^2).
}$$
(To make the patterns clearer to see, I have written $\sqrt{1-\rho^2}=\rho^\prime.$)

That should be good enough, but I'll take you a little closer to the end so you can see where this is all going.  We have already found (at $*$) that
$$\operatorname{Cov}(X^2, X^2) = \operatorname{Var}(X^2) = 2.$$
Plugging this value into the preceding expression and comparing it to $(**)$ shows we would like to demonstrate
$$2\rho^2 = \operatorname{Cov}(X^2, Y^2) = 2\rho^2 + 2\rho\rho^\prime \operatorname{Cov}(X^2, XU) + (\rho^\prime)^2 \operatorname{Cov}(X^2, U^2).$$
Evidently, no matter what value $\rho$ might have, the sum of the last two terms needs to be zero.  This insight reminds us that $X$ and $U$ are independent.  Exploit that fact to show that each of the remaining covariances is zero.  Explicitly, prove
$$\operatorname{Cov}(X^2, XU) = 0 = \operatorname{Cov}(X^2, U^2).$$
