Given that the multivariate random variables ($A_1, A_2, A_3$) have a standard Gaussian on $\mathbb{R}^3$, and a matrix be defined as
$$M= \begin{pmatrix} \sqrt2A_1 &A_3 \\ A_3 & \sqrt2A_2 \end{pmatrix}$$.
Let $X = \alpha_1 + \alpha_2$ and $Y = \alpha_1 - \alpha_2$ where $\alpha_1$ and $\alpha_2$ are the eigenvalues of the matrix M with $\alpha_1 \geq \alpha_2$.
How do I find the density function of $\frac{X}{Y}$?
I tried to find the eigenvalues for $M$ using $det(M - \alpha I) = 0$, and I got
$$\alpha_1 = \frac{\sqrt2A_1 + \sqrt2A_2 + \sqrt{2A_1^2-4A_1A_2+2A_2^2+4A_3^2}}{2}$$ $$\alpha_2 = \frac{\sqrt2A_1 + \sqrt2A_2 - \sqrt{2A_1^2-4A_1A_2+2A_2^2+4A_3^2}}{2}$$ $$\frac{X}{Y} = \sqrt\frac{1}{A_1^2-2A_1A_2+A_2^2+2A_3^2}\biggl(A_1+A_2\biggr)$$
How do I derive the density from here?
I assume $A_1,A_2,A_3$ are independent standard normal.
Going by your work,
$$\frac{X}{Y}=\frac{A_1+A_2}{\sqrt{(A_1-A_2)^2+2A_3^2}}=\frac{(A_1+A_2)/\sqrt 2}{\sqrt{\left(\frac{A_1-A_2}{\sqrt 2}\right)^2+A_3^2}}$$
Change variables $\begin{pmatrix}A_1 \\ A_2 \\ A_3\end{pmatrix}\mapsto \begin{pmatrix}U \\ V \\ W\end{pmatrix}$ such that
$$\begin{pmatrix}U \\ V \\ W\end{pmatrix}= \underbrace{\begin{pmatrix}\frac1{\sqrt 2} & \frac1{\sqrt 2}&0 \\ \frac1{\sqrt 2} & -\frac1{\sqrt 2}&0 \\ 0&0&1\end{pmatrix}}_B \begin{pmatrix} A_1 \\ A_2 \\ A_3\end{pmatrix}$$
Note that $B$ is an orthogonal matrix, which means
$$\begin{pmatrix} A_1 \\ A_2 \\ A_3\end{pmatrix}\sim N_3(\mathbf 0,I_3)\implies \begin{pmatrix} U \\ V \\ W\end{pmatrix} \sim N_3(\mathbf 0,BB^T=I_3), $$
In other words, $U,V,W$ are also independent standard normal.
It is easy to see that $U=\frac{A_1+A_2}{\sqrt 2}, V=\frac{A_1-A_2}{\sqrt 2}$ are standard normal, and that both of them are independent of $W=A_3$. The key is to prove the independence of $U$ and $V$. So if you are not comfortable with the orthogonal transformation above, argue that $U$ and $V$ are jointly normal and that they are uncorrelated. This would imply their independence.
Now $V^2,W^2$ are independent $\chi^2_1$ variables, so that $V^2+W^2 \sim \chi^2_2$, which is again independent of $U$. Thus, $T=\frac{U}{\sqrt{(V^2+W^2)/2}}$ has a $t$ distribution with $2$ degrees of freedom.
And of course,
$$\frac{X}{Y}=\frac{U }{\sqrt{V^2+W^2}}=\frac{T}{\sqrt 2} $$