Check that $f(x,y)=y^{-1}e^{-y-x/y}$ is a joint density function on $x>0$, $y>0$ 
Consider
  $$f(x,y)=\frac{1}{y}e^{-\left(y+\frac{x}{y}\right)} , \qquad 0<x,y<\infty$$
  Verify that $f$ is a joint density function.

$$\int^{y=\infty}_{y=0} \int^{x=\infty}_{x=0}\frac{1}{y} e^{-\left(y+\frac x y \right)} \, dx \, dy$$
$$=-\int^{y=\infty}_{y=0} \left. e^{-\left(y+\frac x y \right)} \right|^\infty_0 \, dy$$
$$=\int^{y=\infty}_{y=0}1 \, dy$$
This does not become $1$. How do I prove it is a joint density function, then?

First there's missing minus sign: you need
$$
\left. -e^{-\left( y + \frac x y \right)} \right|_0^\infty
$$
Instead of
$$
\left. -e^{-\left(y+\frac x y \right)} \right|^\infty_0
$$
it can be useful to write
$$
\left. -e^{-\left(y+\frac x y \right)} \right|^\infty_{x=0},
$$
reminding you which variable goes from $0$ to $\infty$.  The $y$ is still a free variable in this expression.  So you get
$$
\Big( \lim_{x\to\infty} e^{-\left( y + \frac x y \right)} \Big)- \Big(-e^{-\left( y + \frac 0 y \right)} \Big)
$$
which is $0 - \Big( -e^{-y}\Big),$ or $e^{-y}.$
