<s>[INST] Solve $y^{2/3}+(y')^{2/3}=1$ other than the direct method? Is there any way to solve
$$y^{2/3}+(y')^{2/3}=1$$
other than just solving for $y'$ and then integrate?
 [/INST] $\cos^2 u + \sin^2 u = 1 \Rightarrow (\cos^3 u)^{2/3} + (\sin^3 u)^{2/3} = 1$
Let $y = \cos^3 u$ such that $y' = \sin^3 u$, if possible.
$y = \cos^3 u \Rightarrow\\
y' = -3 (\cos^2 u \sin u) u' \Rightarrow\\
-3 u'\cos^2 u \sin u = \sin^3 u \Rightarrow\\
3u' = -\tan^2 u \Rightarrow\\
\displaystyle \int -3 \cot^2 u\, du = x + c \Rightarrow\\
3 \displaystyle \int 1 - \csc^2 u\, dy = x + c \Rightarrow\\
3(u + \cot u) = x + c
$
Now, $y = \cos^3 u \Rightarrow u = \cos^{-1} y^{1/3} \Rightarrow \cot u = \dfrac{y^{1/3}}{\sqrt{1 - y^{2/3}}}$
Thus the solution is
$$\boxed{3\left( \cos^{-1} y^{1/3} + \dfrac{y^{1/3}}{\sqrt{1 - y^{2/3}}} \right) = x + c}$$
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