Find the arc length of the graph of $\displaystyle \large x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$.
Hint: Use symmetry with respect to the line $y=x$.
Let $y=x$ intersect at $a$. So, $\displaystyle a=\left(\frac{1}{2}\right)^{\large \frac{3}{2}}.$
Then $\displaystyle y'=-\frac{\sqrt{1-x^{\large\frac{2}{3}}}}{x^{\large \frac{1}{3}}}$
The arc length is $$\displaystyle \sqrt{1+\frac{\left(1-x^{\large \frac{2}{3}}\right)}{\left(x^{\large \frac{1}{3}}\right)^2}}=x^{-\large \frac{1}{3}}$$
Now, I don't know how to finish. Any suggestions?
Suggestion: For the first quadrant part of the curve, use the parametrization $x=\cos^3\theta$, $y=\sin^3\theta$. Then we want to integrate $$\sqrt{\left(\frac{dx}{d\theta}\right)^2+ \left(\frac{dy}{d\theta}\right)^2},$$ which simplifies very nicely.
Added: For the full arclength using your approach, integrate $x^{-1/3}$ from $0$ to $1$, and multiply by $4$ to take care of the pieces of the curve that lie in the other quadrants. By the symmetry of the curve, these all have the same length as the first quadrant part.
If you want, you can exploit the symmetry between $x$ and $y$ and find the arclength of the first quadrant part by integrating from $0$ to $a$, and multiplying the result by $2$. Then multiply by $4$ to get the full arclength.