For real $u,v$, let
\begin{equation*}
Q(u,v):=P(Y_1>u\sqrt2,Y_2>v\sqrt2)=\int_u^\infty dz\,\varphi(z)\Big(1-\Phi\Big(\frac{2v-z}{\sqrt3}\Big)\Big),
\end{equation*}
where $\varphi$ and $\Phi$ are the standard normal pdf and cdf, respectively.

The question in the OP can be restated as follows: show that
\begin{equation*}
r(u,v):=r_t(u,v):=\frac{Q(u-t,v-t)}{Q(u,v)} \tag{10}\label{10}
\end{equation*}
is increasing in real $u$ and in real $v$ for each $t\in(0,1/\sqrt2)$. Since $u$ and $v$ are interchangeable, it is enough to shown that
\begin{equation*}
r(v):=r(u,v)
\end{equation*}
is increasing in real $v$ for each $t\in(0,1)$.

This will be done by using so-called l'Hospital-type rules for monotonicity. Indeed, consider the "derivative ratio" for the ratio in \eqref{10} with the extra constant factor $e^{t^2/2}$:
\begin{equation*}
r_1(v):=e^{t^2/2}\frac{Q_v(u-t,v-t)}{Q_v(u,v)},
\end{equation*}
where the subscript $_v$ denotes the partial derivative in $v$. We have
\begin{equation*}
Q_v(u,v)=-\int_u^\infty dz\,\varphi(z)\,\frac2{\sqrt3}\,\varphi\Big(\frac{2v-z}{\sqrt3}\Big)
=-\frac{\varphi(v)}2\, \left(\text{erf}\left(\frac{v-2 u}{\sqrt{6}}\right)+1\right).
\end{equation*}
So,
\begin{equation*}
r_1(v)=\frac{f_1(v)}{g_1(v)}, \tag{20}\label{20}
\end{equation*}
where
\begin{equation*}
f_1(v):=e^{t v} \left(\text{erf}\left(\frac{t-2 u+v}{\sqrt{6}}\right)+1\right),\quad
g_1(v):=\text{erf}\left(\frac{v-2 u}{\sqrt{6}}\right)+1.
\end{equation*}
Next, consider the "derivative ratio" for the ratio in \eqref{20}:
\begin{equation*}
r_2(v):=\frac{f'_1(v)}{g'_1(v)}=\frac{f_2(v)}{g_2(v)}, \tag{30}\label{30}
\end{equation*}
where
\begin{equation*}
f_2(v):=\sqrt{6 \pi } t \left(\text{erf}\left(\frac{t-2 u+v}{\sqrt{6}}\right)+1\right)+2
e^{-(t-2 u+v)^2/6},\quad
g_2(v):=2 e^{-t v-(v-2 u)^2/6}.
\end{equation*}
Further, consider the third (and final) "derivative ratio" for the ratio in \eqref{30}:
\begin{equation*}
r_3(v):=\frac{f'_2(v)}{g'_2(v)}=\frac{e^{-t (t-4 (u+v))/6}\, (-2 t-2 u+v)}{3 t-2 u+v}. \tag{40}\label{40}
\end{equation*}
Note that for all real $z$
\begin{equation}
r'_3(2u+z)=\frac{t \left(2 z^2+2 t z+15-12 t^2\right) e^{-t (t-4 (3 u+z))/6}}{3 (3
t+z)^2},
\end{equation}
which is obviously $>0$ for all $t\in(0,1)$ if $z\ne-3t$.

So, $r_3$ is increasing on $(-\infty,2u-3t)$ and on $(2u-3t,\infty)$. Also, $g_2>0$. Also, $g'_2>0$ on $(-\infty,2u-3t)$ and $g'_2<0$ on $(2u-3t,\infty)$.

So, by lines 1 and 3 of Table 1.1, (the continuous function) $r_2$ is down-up on $(-\infty,2u-3t]$ and up-down on $[2u-3t,\infty)$ -- that is, (i) there is some $a\in[-\infty,2u-3t]$ such that $r_2$ is decreasing on $(-\infty,a]$ and increasing on $[a,2u-3t]$ and (ii) there is some $b\in[2u-3t,\infty]$ such that $r_2$ is increasing on $[2u-3t,b]$ and decreasing on $[b,\infty)$.

So, $r_2$ is down-up-down on $(-\infty,\infty)$. But $r_2$ is a positive function such that $r_2(v)\to0$ as $v\to-\infty$ and $r_2(v)\to\infty$ as $v\to\infty$. So, $r_2$ is increasing on $(-\infty,\infty)$.

But $f_1(v)\to0$ and $g_1(v)\to0$ as $v\to-\infty$. So, by Proposition 4.1 of the cited paper, $r_1$ is increasing on $(-\infty,\infty)$.

Also, clearly $Q(u,v)\to0$ as $v\to\infty$. So, again by Proposition 4.1 of the cited paper, $r(u,v)$ is increasing in $v\in (-\infty,\infty)$. $\quad\Box$