Let me rephrase the question in a somewhat more orthodox language. One considers two random variables $X=uZ$ and $Y=vZ$, where $Z$ is standard normal and $u$ and $v$ are real numbers such that $(u,v)\ne(0,0)$.
Then $(X,Y)\in D$ almost surely, where $D$ is the straight line $D=\{(x,y)\in\mathbb R^2\mid vx=uy\}$. Since $D$ has Lebesgue measure zero, the distribution of $(X,Y)$ has no density.
Note that all that matters is that one can still evaluate expectations, using the density $f_Z$ of the random variable $Z$. To wit, for every measurable function $g$, $ \mathbb E(g(X,Y))=\int_{-\infty}^{+\infty}g(uz,vz)\,f_Z(z)\,\mathrm dz. $
Edit: The fact that $D$ has Lebesgue measure zero is not a probability result since the Lebesgue measure on the plane has infinite mass but here is a proof. Consider without loss of generality the line $\Delta=\mathbb R\times\{0\}$, then $\Delta=\bigcup\limits_{n\geqslant1}\Delta_n$ with $\Delta_n=[-n,n]\times\{0\}$. Note that, for every $\varepsilon\gt0$, $\Delta_n\subset[-n,n]\times[-\varepsilon,\varepsilon]$. The rectangle $[-n,n]\times[-\varepsilon,\varepsilon]$ has Lebesue measure $4n\varepsilon$, hence the Lebesgue measure of $\Delta_n$ is at most $4n\varepsilon$, for every $\varepsilon\gt0$, hence the Lebesgue measure of $\Delta_n$ is zero, hence the Lebesgue measure of their union $\Delta$ is zero.
Edit: See Elementary Probability for Applications, by Rick Durrett.