First, the symmetries of the curve $xy = \nu$.
The equation is unchanged by $y' = -x$ and $x' = -y$ leading to the line of symmetry $y = -x$.
The equation is symmetric with respect to $x$ and $y$ showing that there is a line of symmetry $y = x$.
I tried to find the line of symmetry of a more complicated curve such as $x^5 + y^3 + xy = 5$.
I could not apply my strategy shown above because it relied on lucky spots.
I tried to find a way to find the lines of symmetry of any curve I could come up with algebraically.
I could not find a way.
Is there a way?
I thought that if I have, just for the above example, my curve $f(x, y) = const$ then the curve's symmetry is where $f(x', y') = f(x, y)$. This means I would need new variables $x'$ and $y'$ such that
$$x'^5 + y'^3 + x'y' = x^5 + y^3 + xy = 5$$
Obviously the question is then how do I find those variables. I could work on the assumption that they are related to the originals by something like $x' = x + a, y' = y + b$ - I'll try that now.
Using this I have
$$(x + a)^5 + (y + b)^3 + (x + a)(y + b) = x^5 + y^3 + xy$$
Giving
$$x^5 + 5x^4a + 10x^3a^2 + 10x^2a^3 + 5xa^4 + a^5 + y^3 + 3y^2b + 3yb^2 + b^3 + xy + xb ya + ab = x^5 + y^3 + xy$$
Now I can compare coefficients in x and y, although this seems to just suggest that a = b = 0! This cannot be right.