I'm trying to find the arc length of the curve defined by $$(x-1)^{2/3}+(y-2)^{2/3}=1$$.My first approach was try to set 'y' in terms of 'x' and then apply the formula $$L=\int\limits_a^b\sqrt{1+y'}dx$$ But, this approach is too difficult regarding we do not know $a$ or $b$ and the result expression certainly is not simple.
Then I tried to parametrize the curve, but I can't find any good parametrization in order to make the integral as simple as possible. If you know any good method to find arc lengths of curves defined by expressions with the form $$F(x,y)=0$$ that'd be really helpful. Thanks.
Use the parameterization
$$x=1+\cos^3 t, \quad y=2+\sin^3 t$$
for $0\le t\le 2\pi$. The curve is symmetric, so you can find four times the length of the curve for $0\le t\le \pi/2$. (In this problem that avoids some sign issues and makes it simpler.) The length of a parameterized curve is
$$\int_{t=a}^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$$
so use four times that expression for the given $x$ and $y$, $a=0$, $b=\pi/2$.
The integral is not difficult. Here is a graph, showing the complete curve as you defined it in red, and the parameterized quarter of the curve in cyan. This shows that the parameterization and limits are correct.