Let $0<\theta_1,\theta_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.
Consider the oscillatory integral
$$I(t):=\int_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta_1})
(y-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx dy.$$
I am trying to find the asymptotic behavior of $I(t)$ as $t\rightarrow \infty$.
Since the phase $x\mapsto t x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives
$$\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx= \frac{C \psi(0)}{\sqrt{t}y}+O(\frac{1}{t y^2}),$$
with some constant $C$. But then we are faced with the singularities $y^{-1}$ and $y^{-2}$.
If we use Fubini's theorem then apply the stationary phase method we end up with
$$\int_{0}^{1}\frac{e^{\dot{\imath} t x^2y^2}}{(y-e^{\dot{\imath}\theta_1})
(y-e^{\dot{\imath}\theta_2})}dr= \frac{\widetilde{C} }{\sqrt{t}x}+O(\frac{1}{t x^2}),$$
which is again not integrable against $\psi$.
Another approach is to change variables $s=\sqrt{t}r$ to write
$$I(t)=\frac{1}{\sqrt{t}}
J(t)$$
where
$$J(t):=\int_{0}^{\sqrt{t}}\frac{1}{(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_1})
(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta_2})}\int_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$
The idea now is to check whether $\lim_{t\rightarrow \infty}J(t)$
is a constant independent of $t$. But, by Fubini's theorem, we have
$$\lim_{t\rightarrow \infty}J(t)=c\int_{\mathbb{R}}
\frac{\psi(x)}{x}dx$$ which may diverge.
Best Answer
$\newcommand{\R}{\mathbb R}\newcommand\sgn{\operatorname{sgn}}\newcommand{\vpi}{\varphi}$Obviously, for $a:=\sqrt{2\pi}\,\psi(0)$ we have \begin{equation*} \psi(0)=f(0), \end{equation*} where \begin{equation*} f:=a\vpi \end{equation*} and $\vpi$ is the standard normal density. Letting now $g(x):=\frac{\psi(x)-f(x)}x$ for $x\ne0$, with $g(0):=\psi'(0)$, we see that $g$ is a smooth integrable function and \begin{equation*} \psi(x)=f(x)+xg(x) \tag{1}\label{1} \end{equation*} for all real $x$.
So, \begin{equation*} I(t)=\frac{J_f(t)+J_h(t)}{\sqrt t}, \tag{2}\label{2} \end{equation*} where $h(x):=xg(x)$, \begin{equation*} J_f(t):= \int_0^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)} \int_{\R}dx\,f(x)e^{ix^2y^2}, \end{equation*} \begin{equation*} z_1:=e^{i\theta_1},\quad z_2:=e^{i\theta_2}, \end{equation*} with $J_h(t)$ defined similarly. Here and in what follows, $t$ is any real number $>0$, unless specified otherwise.
Note that \begin{equation*} \Im z_1\ne0\quad\text{and}\quad\Im z_2\ne0. \tag{3}\label{3} \end{equation*} Note also that \begin{equation*} \int_{\R}dx\,f(x)e^{ix^2y^2}=\frac a{\sqrt{1-2 i y^2}} \end{equation*} and hence \begin{equation*} \begin{aligned} J_f(t)&= a\int_0^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}}. \end{aligned} \tag{4}\label{4} \end{equation*} For any fixed real $A>0$, by dominated convergence (which holds in view of \eqref{3}), \begin{equation*} \int_0^A\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}} \to\frac1{z_1z_2}\int_0^A\frac{dy}{\sqrt{1-2 i y^2}}\ll1 \end{equation*}(as $t\to\infty$). We write $E\ll F$ to mean $|E|=O(F)$.
So, letting $A$ go to $\infty$ slowly enough, we will have $A=o(\sqrt t)$ and \begin{equation*} \int_0^A\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}} =o(\ln t). \end{equation*} Also, we will have \begin{equation*} \int_A^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}} \\ \sim\int_A^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)y\sqrt{-2 i}} \sim\frac{1+i}{4z_1z_2}\,\ln t. \tag{5}\label{5} \end{equation*} The latter asymptotic expression in \eqref{5} can be obtained by taking the latter integral in \eqref{5} in closed form, by partial fraction decomposition. It can also be obtained by writing $\int_A^{\sqrt t}=\int_A^{t^b}+\int_{t^b}^{t^{1/2}}$ for $b\in(0,1/2)$ such that $b$ is close to $1/2$.
So, assuming $\psi(0)\ne0$, by \eqref{4}, \begin{equation*} J_f(t)\sim a\frac{1+i}{4z_1z_2}\,\ln t =\frac{1+i}{4z_1z_2}\,\sqrt{2\pi}\,\psi(0)\ln t. \tag{6}\label{6} \end{equation*}
Next, \begin{equation*} J_h(t)= \int_{\R}dx\,\sgn(x)g(x)K(t,x), \tag{7}\label{7} \end{equation*} where \begin{equation*} K(t,x):=\int_0^{\sqrt t}\frac{dy\,e^{ix^2y^2}|x|}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)} =\frac12\int_0^{r^2}dv\, e^{iv}H_r(v), \end{equation*} \begin{equation*} r:=|x|\sqrt t, \end{equation*} \begin{equation*} H_r(v):=\frac1{\sqrt v\,(\frac{\sqrt v}r-z_1)(\frac{\sqrt v}r-z_2)}. \end{equation*} Further, \begin{equation*} 2K(t,x)=K_1+K_2, \tag{8}\label{8} \end{equation*} where \begin{equation*} K_1:=\int_0^{1\wedge r^2}dv\, e^{iv}H_r(v),\quad K_2:=\int_{1\wedge r^2}^{r^2}dv\, e^{iv}H_r(v), \end{equation*} $u\wedge w:=\min(u,w)$.
For $v\in(0,1\wedge r^2)$, we have $H_r(v)\ll\frac1{\sqrt v}$ and hence \begin{equation*} K_1\ll1. \tag{9}\label{9} \end{equation*} Note that $K_2=0$ if $r\le1$. If now $r>1$, then \begin{equation*} H'_r(v)=-\frac{H_r(v)}{2v}\,\Big(1+\frac1{\sqrt v-rz_1}+\frac1{\sqrt v-rz_2}\Big) \ll \frac{|H_r(v)|}v\ll\frac1{v^{3/2}} \end{equation*} for $v>0$; so, integrating by parts, we see that
\begin{equation*} K_2\ll1. \tag{10}\label{10} \end{equation*}
Collecting \eqref{7}--\eqref{10} and recalling that $g$ is an integrable function, we see that \begin{equation*} J_h(t)\ll1. \tag{11}\label{11} \end{equation*}
Collecting \eqref{2}, \eqref{6}, and \eqref{10}, we conclude that \begin{equation*} I(t)\sim \frac{1+i}{4z_1z_2}\,\sqrt{2\pi}\,\psi(0)\frac{\ln t}{\sqrt t}, \tag{12}\label{12} \end{equation*} as $t\to\infty$.
As seen from the above proof, for \eqref{12} to hold it is enough that the function $\R\setminus\{0\}\ni x\mapsto\frac{\psi(x)-\psi(0)}x$ be integrable, with $\psi(0)\ne0$. (Of course, $g$ will be integrable if, as stated by the OP, "$\psi$ is a smooth real-valued function with compact support".)
Concerning the case $\psi(0)=0$: Then the bounds on $H_r(v)$ and $H'_r(v)$ developed above provide for dominated convergence. So, taking into account that for each real $x\ne0$ \begin{equation} H_r(v)\to\frac1{z_1z_2\sqrt v}, \end{equation} we have \begin{equation} K(t,x)\to\frac1{2z_1z_2}\int_0^\infty\frac{dv\,e^{iv}}{\sqrt v} =\frac{1+i}{2z_1z_2}\,\sqrt{\frac\pi2} \end{equation} and \begin{equation*} J_h(t)\to C_\psi:= c_\psi \frac{1+i}{2z_1z_2}\,\sqrt{\frac\pi2}, \end{equation*} where \begin{equation} c_\psi:=\int_{\R}dx\,\sgn(x)g(x)=\int_{\R}dx\,\frac{\psi(x)}{|x|}. \end{equation} Therefore and because here $f=0$ and hence $J_f(t)=0$, we conclude that \begin{equation*} I(t)\sim \frac{C_\psi}{\sqrt t}, \end{equation*} as $t\to\infty$, provided that $\psi(0)=0$, $c_\psi\ne0$, and the function $g$ is integrable. (In the case when $\psi(0)=0$, we have $g(x)=\psi(x)/x$ for $x\ne0$. As was already noted, $g$ will be integrable if, as stated by the OP, "$\psi$ is a smooth real-valued function with compact support".)
If $\psi(0)=0$ and $c_\psi=0$, then one has to dig deeper yet, possibly ad infinitum.
However, if $\psi$ is a "smooth bump", as you said in a comment, then apparently $\psi\ge0$ and hence $c_\psi>0$ (unless $\psi=0$ and hence $I(t)=0$).