# Find the PDF of G(X,Y)=$(X,\frac{9}{(X+Y)^{2}})$ Transformation given the pdf of $f_{XY}(x,y)$

So I made up a problem for myself, and am sort of confused on how to get started. if you work out the full solution I will be sad!

This is a deviation from a book problem in Casella and Burger 4.4 where I've made the problem a bit harder. and am stuck.

so here is the joint pdf $f(x,y)=\frac{1}{4}(x+2y)$ for $0<y<1$ and $0<x<2$ and 0 other wise.

My question has to do with transformations of multi-variable probability densities Let $G(X,Y)=(g_{1}(X,Y),g_{2}(X,Y))= (X,\frac{9}{(X+Y)^{2}})$ be a random variable $(X,Y) \to (X,V)$

Now assume that G is is invertible and now we find the inverse on this interval.

doing some algebra returns H(U,V)= (U,$\frac{3}{\sqrt{V}}-U)$. a little more algebra to find the Jacobian returns $|J| = \frac{3}{2}v^{-\frac{3}{2}}$ Now finding the joint pdf under the transformation $$\begin{split} f_{UV}(u,v) &=\frac{1}{4}(u + 2(\frac{3}{\sqrt{v}} - u))\cdot \frac{3}{2}v^{-\frac{3}{2}} \\ &= \frac{3}{8}(\frac{6}{v^{2}}- \frac{u}{v^{\frac{3}{2}}}) \end{split}$$

now assuming I escaped an algebra mistake, is the set up that I've done correct? essentially the conceptual idea is that given a joint PDF of X,Y I looked at a transformation of G, found the inverse and the Jacobian. and now the joint pdf $f_{UV}(u,v)$ is joint pdf of X,Y under the map of G ?
thank you all for any help

What you have now is correct, but technically to complete the joint pdf of $U$ and $V$ you need to include their support.
Note that $$0 < x < 2 \iff 0 < u < 2$$ and $$0 < y < 1 \iff 0 < \frac{3}{\sqrt{v}} - u < 1 \iff \cdots \iff \frac{9}{(u+1)^2} < v < \frac{9}{u^2}$$
and you can check that $$\int_0^2 \int_{9/(u+1)^2}^{9/u^2} \frac{3}{8}\left(\frac{6}{v^{2}}- \frac{u}{v^{\frac{3}{2}}} \right) \; dv \, du =1.$$ So the pdf is $$f_{U,V}(u,v) = \begin{cases} \frac{3}{8}\left(\frac{6}{v^{2}}- \frac{u}{v^{\frac{3}{2}}} \right), & 0 < u < 2 \text{ and } \frac{9}{(u+1)^2} < v < \frac{9}{u^2}, \\ 0, & \text{else.} \end{cases}$$