Let X and Y be normal random variables with mean 0, variance $\sigma^2$ and correlation coefficient $\rho \in (-1,1)$, so that the density is given by
$$f(x,y) = \cfrac{1}{2\pi \sigma^2\sqrt{a-\rho^2}}\,\,\exp{\{-\cfrac{1}{2\sigma^2(1-\rho^2)}[x^2 - 2\rho xy+y^2]\}}$$
How to determine the distribution of $Z=\cfrac{X}{Y}$ ?
Let $Z=X/Y$ and $W=Y$.
The transformation $z=x/y,\: w=y$ has the inverse transformation
$x=zw, \: y=w$, and
$$ \bar{J}(z,w) =\begin{vmatrix} \frac{\partial{x}}{\partial{z}} & \frac{\partial{x}}{\partial{w}} \\ \frac{\partial{y}}{\partial{z}} & \frac{\partial{y}}{\partial{w}} \\ \end{vmatrix} $$
$$ \bar{J}(z,w)=\begin{vmatrix} w & z \\ 0 & 1 \\ \end{vmatrix} = w $$
Now
$$f_{ZW}(z,w) = |w|f_{XY}(zw,w)$$
and the marginal pdf of $Z$ is
$$f_{Z}(z) = \int_{-\infty}^{\infty}|w|f_{XY}(zw,w)dw$$