I am working on prep questions for the Qualifying exam and stuck with the following one. I tried to find the joint density using the transformation by the Jacobian, but it did not work out because I could not figure out the inverse of $U$ and $V$. Any helps would be appreciated!
Let $X$ and $Y$ be independent Exponential random variables with mean 1. Let $$U = \exp(X) + 2\exp(Y), V = 2\exp(X^2) + \exp(Y^2)$$ and let $g(u,v)$ be a joint density for $U$ and $V$ which is as continuous as possible. Find the value of $g(3e,3e)$.
Define $(u,v)=\big(e^x+2e^y,2e^{x^2}+e^{y^2}\big)$ where $(x,y)\in (0,\infty)^2$. Observe how $$(u,v)=(3e,3e) \iff (x,y)=(1,1)$$ Therefore $$g(3e,3e)=f_{XY}(1,1)\Bigg|\frac{\partial(x,y)}{\partial(u,v)}\Bigg|_{(u,v)=(3e,3e)}\Bigg|=\frac{f_{XY}(1,1)}{\Bigg|\frac{\partial(u,v)}{\partial(x,y)}\Bigg|_{(x,y)=(1,1)}\Bigg|}$$ Since $X,Y\sim \exp(1)$ are independent we have $f_{X,Y}(x,y)=f_{X}(x)f_{Y}(y)=e^{-(x+y)}$ which is supported $(0,\infty)^2$. Also $\frac{\partial(u,v)}{\partial(x,y)}=2ye^xe^{y^2}-8xe^{x^2}e^y$ so that $g(3e,3e)=\frac{e^{-2}}{\big|-6e^2\big|}=\frac{1}{6e^4}$.