Let the function $f : \mathbb{R} \rightarrow \mathbb{R}$ be differentiable and its derivative is never zero. Show that these curves are orthogonal:
$$f(xy) = C$$ $$ y^2 -x^2 = D$$
$C$ and $D$ are real numbers.
Edit: I checked the duplicate flag and I do not fully agree with it, for this question is somewhat different. Furthermore, after I posted (as my first post) this question I solved it and was writing an answer for it. It took some time because of the formatting. But then the post closed which made me lose my progress.
In conclusion, to the curators who are reading this, do not dismiss questions so easily.
Anyway, where is my answer:
Let $$ g(x,y) = xy $$ $$ f\circ g = f(g(x,y)) = f(xy)$$
Differentiating with respect to $x$ the first two equations we get:
$$ (f(xy))' = 0 $$ $$ 2yy' -2x = 0 $$
Applying the chain rule, and simplifying we get:
$$ (y+xy')f'(xy) = 0 $$ $$ y' = \frac{x}{y} $$
Simplifying further the first equation (we can divide by $f'(xy)$ because its never zero) we get:
$$ y' = -\frac{y}{x} = m_1 $$ $$ y' = \frac{x}{y} = m_2$$
Since $ m_1 = -\frac{1}{m_2} $, we can assume that the tangent lines of the first family of curves is perpendicular to the tangent lines of the second family of curves. Making these curves orthogonal.
Edit 2: The duplicate is more alike than I thought, the curator is most likely correct.