<s>[INST] Smooth parametrization of curve $\cosh x+\cosh y = \operatorname{constant}$ I have a curve (image below)
$$
\cosh x+\cosh y = C,\qquad C>2.
$$
I would like to get its smooth parametrization of form
$$
x = f(t),\qquad y=g(t),\qquad t\in[a,b],
$$
so for every point on the curve there is a corresponding parameter $t$. (In the same manner, as for curve $x^2+y^2=1$, there is a smooth parametrization $x=\cos t$, $y = \sin t$).
I would appreciate any help. Thanks!

 [/INST] Here's something more symmetric. I'll write the steps if anyone wants to modify the idea to get something else.
First, we can remove the pesky hyperbolic functions simply by letting $x = \ln t,\ y = \ln s$ so $$\cosh x + \cosh y = c \iff t + \frac 1t + s + \frac 1s = 2c.$$
Now, we'd like to "invert" $z\mapsto z + 1/z$. More precisely, we want $t + 1/t = 2u^2+2$ and $s + 1/s = 2v^2 + 2$, so the whole thing becomes $ u^2 + v^2 = c - 2$ which we know how to parametrize.
Solving the appropriate quadratics, we get $$t = 1 + u^2\pm \sqrt{2u^2+u^4},\ s = 1+v^2 \pm \sqrt{2v^2 + v^4}.$$ Noting that we want $ u^2 + v^2 = c - 2$, we further choose $u = \sqrt {c-2} \cos\varphi$ and $v = \sqrt {c-2}\sin\varphi$.
All in all, $$x = \ln \left( 1 + (c-2)\cos^2\varphi \pm \sqrt{2(c-2)\cos^2\varphi + (c-2)^2\cos^4\varphi} \right),\\
y = \ln \left( 1 + (c-2)\sin^2\varphi \pm \sqrt{2(c-2)\sin^2\varphi + (c-2)^2\sin^4\varphi} \right).
$$
Now you must be wondering what to do with these $\pm$ signs. Well, if you choose one of the four possibilities, you will get the part of the curve in the corresponding quadrant. So, to deal with that:
$$x = \ln \left( 1 + (c-2)\cos^2\varphi + \cos\varphi \sqrt{2(c-2) + (c-2)^2\cos^2\varphi} \right),\\
y = \ln \left( 1 + (c-2)\sin^2\varphi + \sin\varphi \sqrt{2(c-2) + (c-2)^2\sin^2\varphi} \right),
$$ and we are finally done.
</s>