<s>[INST] To prove $X^2$ and $Y^2$ are independent for a given joint pdf $f(x,y)=\dfrac{1}{4}(1+xy)$ To prove $X^2$ and $Y^2$ are independent for a given joint pdf $f(x,y)=\dfrac{1}{4}(1+xy)$
$f(x,y) =
\begin{cases}
\dfrac{1}{4}(1+xy),  & |x|<1,|y|<1 \\
0, & \text{otherwise}
\end{cases}$
The given solution starts with trying to prove that 
\begin{align}
$P(X^2\leq x \cap Y^2\leq y) & = P(|X|\leq \sqrt{x} \cap |Y|\leq \sqrt{y}) \\
 & = \int\limits_{-\sqrt{x}}^{\sqrt{x}}\int\limits_{-\sqrt{y}}^{\sqrt{y}}f(u,v)\,du\,dv \\ 
 & = \int\limits_{-\sqrt{x}}^{\sqrt{x}}\int\limits_{-\sqrt{y}}^{\sqrt{y}}\dfrac{1}{4}\,du\,dv
\end{align}
I did not understand how we got to $f(u,v)=\dfrac{1}{4}$ rather that $\dfrac{1}{4}(1+xy)$
I tried with $X^2=u$ and $Y^2=v$, does not work out.
Please help.
 [/INST] The answer is rather simple. Just verify by a simple calculation that
$\frac{1}{4} \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} uv \, du dv = 0$ 
and you're done.
Edit: here all steps: 
$\int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} f(u,v) \, du dv  =  \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} \frac{1}{4} \, du dv +  \frac{1}{4} \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} uv \, du dv =  \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} \frac{1}{4} \, du dv + \int\limits_{-\sqrt{x}}^{-\sqrt{x}} \left[ \frac{1}{2} uv^2 \right] _{-\sqrt{y}}^{-\sqrt{y}} \, du =  \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}}  \frac{1}{4} \, du dv + \int\limits_{-\sqrt{x}}^{-\sqrt{x}} 0 \, du = \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}}  \frac{1}{4} \, du dv$ 
</s>