**HINT**

$\color{brown}{\textbf{Standing of the task.}}$

Denote
$$u=t_2,\quad v=t_1,$$
$$p = \log\dfrac{u}{x},\quad q = \log\dfrac{v}{y},\quad r = \log\dfrac{1-u}{1-x},\quad s = \log\dfrac{1-v}{1-y},\quad w = \log\dfrac{v-u}{y-x},\quad $$
$$f(u,v) = s(w-r)(q-p) - q(w-p)(s-r) = sr(p-q) - swp + pq(s-r) +qwr,$$
$$f(u,v) = \begin{vmatrix}w-p-s & w-q-r \\ ps & rq\end{vmatrix},$$
then
\begin{align}
&f_{mn}(u,v) = \dfrac{\partial^{m+n}f_{mn}(u,v)}{\partial u^m\partial v^n}\\[4pt]
&= \sum\limits_{i=0}^m\sum\limits_{j=0}^n \binom mi \binom nj
\begin{vmatrix}
w_{ij}-\delta_{0j}\,p_i-\delta_{i0}\,s_{j} & w_{ij}-\delta_{0j}\,q_i-\delta_{i0}\,r_j \\
p_{m-i}\,s_{n-j} & r_{m-i}\,q_{n-j}
\end{vmatrix},\quad\text{where}\\
&p_i = p^{(i)}(u) = \delta_{i0}\log\dfrac{u}{x} + (\delta_{i0}-1)^{i-1}\,\dfrac{(i-1)!}{u^i},\\
&q_j = q^{(j)}(v) = \delta_{0j}\log\dfrac{v}{y} + (\delta_{0j}-1)^{j-1}\,\dfrac{(j-1)!}{v^j},\\
&r_i= r^{(i)}(u) = \delta_{i0}\log\dfrac{1-u}{1-x} + (\delta_{i0}-1)\,\dfrac{(i-1)!}{(1-u)^i},\\
&s_{ij} = s^{(j)}(v)
=\delta_{0j}\log\dfrac{1-v}{1-y} + (\delta_{0j}-1)\,\dfrac{(j-1)!}{(1-v)^j},\\
&w_{ij}=\dfrac{\partial^{i+j}w}{\partial u^i\partial v^j}
= \delta_{i0}\delta_{0j}\log\dfrac{v-u}{y-x} + (\delta_{i0}\delta_{0j}-1)^{\delta_{i0}+j-1}\,\dfrac{(i-1)!(j-1)!}{(v-u)^{i+j}},\\[4pt]
\end{align}

$\color{brown}{\textbf{Derivatives in the given point.}}$

Note that
\begin{align}
&p(x) = 0,\quad r(x) = 0,\quad q(y) = 0,\quad s(y) = 0,\quad w(x,y) = 0,
\quad f(x,v) = f(u,y) = 0,\\[4pt]
&p_i(x) = (\delta_{i0}-1)^{i-1}\,\dfrac{(i-1)!}{x^i},\\
&q_j(y) = (\delta_{0j}-1)^{j-1}\,\dfrac{(j-1)!}{y^j},\\
&r_i(x) = (\delta_{i0}-1)\,\dfrac{(i-1)!}{(1-x)^i},\\
&s_j(y) = (\delta_{0j}-1)\,\dfrac{(j-1)!}{(1-y)^j},\\
&w_{ij}(x,y) = (\delta_{i0}\delta_{0j}-1)^{\delta_{i0}+j-1}\,\dfrac{(i-1)!(j-1)!}{(y-x)^{i+j}}\\[4pt]
&=\begin{cases}
0,\quad\text{if}\quad i=0\wedge j=0\\
(-1)^j\,\dfrac{(j-1)!}{(y-x)^j},\quad\text{if}\quad i=0 \wedge j>0\\
-\dfrac{(i-1)!}{(y-x)^i},\quad\text{if}\quad i>0 \wedge j=0\\
(-1)^{j-1}\,\dfrac{(i-1)!(j-1)!}{(y-x)^{i+j}},\quad\text{if}\quad i>0 \wedge j>0,\\
\end{cases}\\[4pt]
&f_{mn}(x,y) = \sum\limits_{i=1}^m\sum\limits_{j=1}^n \binom mi \binom nj
\begin{vmatrix}
a_{ij}(x,y) & b_{ij}(x,y) \\ p_{m-i}(x)s_{n-j}(y) & r_{m-i}(x)q_{n-j}(y)
\end{vmatrix},\\
&a_{ij}(x,y) = w_{ij}(x,y)-\delta_{0j}p_i(x)-\delta_{i0}s_j(y),\\
& b_{ij}(x,y) = w_{ij}(x,y)-\delta_{0j}r_i(x)-\delta_{i0}q_j(y),\\
\end{align}
Taking in account zero values of $p,q,r,s,w$ in the given point $\dbinom uv = \dbinom xy,$ unzero derivatives can start from $i+j=3.$

$\color{brown}{\textbf{2D Taylor series.}}$

The previous results allow to present the issue equality in the form of $f_l(u,v)=0,$
where
$$f_l(u,v) = \sum\limits_{k=3}^l \dfrac1{k!}\sum\limits_{m=1}^{k-1}\binom km f_{m,k-m}(x,y)(u-x)^m(v-y)^{k-m}.$$
is $2D$ Taylor series of $l-$th order.

The calculated values of the derivatives are in the table below.

\begin{vmatrix}
& i=0 & i = 1 & i=2 & i = 3 \\
p_i(x) & 0 & \dfrac1x & -\dfrac1{x^2} & \dfrac2{x^3} \\
q_i(y) & 0 & \dfrac1y & -\dfrac1{y^2} & \dfrac2{y^3} \\
r_i(x) & 0 & -\dfrac1{1-x} & -\dfrac1{(1-x)^2} & -\dfrac2{(1-x)^3}\\
s_i(y) & 0 & -\dfrac1{1-y} & -\dfrac1{(1-y)^2} & -\dfrac2{(1-y)^3} \\
w_{i0}(x,y) & 0 & -\dfrac1{y-x} & -\dfrac1{(y-x)^2} & -\dfrac2{(y-x)^3} \\
w_{i1}(x,y) & \dfrac1{y-x} & \dfrac1{(y-x)^2} & \dfrac2{(y-x)^3} & \dfrac2{(y-x)^4} \\
w_{i2}(x,y) & -\dfrac1{(y-x)^2} & -\dfrac2{(y-x)^3} & -\dfrac6{(y-x)^4} & -\dfrac{24}{(y-x)^5} \\
\end{vmatrix}

$\color{brown}{\textbf{The third order model.}}$

One can get
\begin{align}
&a_{01} = w_{01}(x,y)-p_0(x)-s_1(y) = \dfrac1{y-x} + \dfrac1{1-y}=\dfrac{1-x}{(y-x)(1-y)},\\
&b_{01} = w_{01}(x,y)-r_0(x)-q_1(y) = \dfrac1{y-x}-\dfrac1y = -\dfrac x{y(y-x)},\\
&a_{10} = w_{01}(x,y)-p_1(x)-s_0(y) = -\dfrac1{y-x} - \dfrac1x = -\dfrac{y}{x(y-x)},\\
&b_{10} = w_{01}(x,y)-r_1(x)-q_0(y) = \dfrac1{y-x}+\dfrac1{1-x} = \dfrac{1+y-2x}{(1-x)(y-x)},\\
&f_{12} = a_{01}r_1(x)q_1(y) - b_{01}p_1(x)s_1(y)
= -\dfrac{a_{01}}{(1-x)y} + \dfrac{b_{01}}{x(1-y)} = -\dfrac2{(y-x)y(1-y)},\\
&f_{21} = a_{10}r_1(x)q_1(y) - b_{10}p_1(x)s_1(y)
= -\dfrac{a_{10}}{(1-x)y} + \dfrac{b_{10}}{x(1-y)}\\
&= \dfrac1{x(y-x)(1-x)} + \dfrac{1+y-2x}{x(1-x)(y-x))(1-y)}
=\dfrac2{x(y-x)(1-y)},\\
&f_3(u,v) = \dfrac12(f_{12}(v-y)+f_{21}(u-x))(u-x)(v-y)
=\left(\dfrac{u-x}{x}-\dfrac{v-y}{y}\right)\dfrac{(u-x)(v-y)}{(y-x)(1-y)},\\
&f_3(u,v)=0\rightarrow (u=x)\vee(v=y)\vee(uy=vx).
\end{align}

The order of model can be increased.