When I came across problem posted by Eyesima, https://math.stackexchange.com/a/3569114/751970 , I realized that for functions which are symmetrical in $x$ and $y$, their tangent line at points $(a,a)$ can be found easily without tedious derivative, especially if the functions are complicated.
This led me to question if similar concept can be used for functions which are asymmetric in $x$ and $y$ and, it can, under some conditions of course.
For example, find the tangent line of the following curve at $(1,1)$.
$$ \tan^{-1}{\left(xy^{2}\sqrt{\frac{2}{xy+y}}\right)}-\frac{\pi}{4}\sqrt{\frac{3}{xy+xy^{2}+y}}=0 $$
Although the curve is asymmetric in $x$ and $y$, it is symmetric in $xy$ and $y$.
$$ \begin{aligned} \tan^{-1}{\left(xy^{2}\sqrt{\frac{2}{xy+y}}\right)}-\frac{\pi}{4}\sqrt{\frac{3}{xy+xy^{2}+y}}&=0\\ \tan^{-1}{\left(xy\cdot y\sqrt{\frac{2}{xy+y}}\right)}-\frac{\pi}{4}\sqrt{\frac{3}{xy+xy\cdot y+y}}&=0 \end{aligned} $$
Furthermore, at $(1,1)$, $xy=y$. Therefore, at $(1,1)$ the following equation applies
$$ \begin{aligned} d(xy)+dy&=0\\ x\ dy + y\ dx + dy&=0\\ dy+dx+dy&=0\\ 2dy+dx&=0 \end{aligned} $$
and the tangent line is $2y+x-3=0$.
This is only one example.