Determining the "perfect" $\delta$ in the uniform continuity of $\sqrt[3]{1-x^2}$ The textbook problem was to find, for every $\varepsilon\gt0$, some $\delta(\varepsilon)$ such that for all $x,y\in[-1,1]:|x-y|\lt\delta,\,|\sqrt[3]{1-x^2}-\sqrt[3]{1-y^2}|\lt\varepsilon$. This problem is not too bad: on my second try, I found that $\delta=\varepsilon^3/2$ works just fine in that interval.
For my own interest, I pursued this further beyond the scope of the question:
On my first try, I used some very inaccurate reasoning to actually get a very good result; I found that $(\varepsilon/2)^3$ works incredibly well even outside that interval, and a visual check with Desmos finds only a tiny, tiny sliver for which that $\delta$ fails.
A week later, I realised my derivation was flawed (at least I believe it is flawed), but after a week further still of some attempts to track down the "perfect" $\delta$ (i.e, the largest, or close to largest, $\delta$ such that the uniform continuity condition is satisfied) I have failed, and was hoping someone could help me with the final step I got stuck on below. I have reason to believe this perfect delta exists, as visually my incorrect answer of $(\varepsilon/2)^3$ is very close, and it does seem possible to maximise $\delta$.
My attempt to get the tightest inequality possible (the working may seem brief but I have tried and failed with many other approaches, this approach was the most succesful):

Let $a=\sqrt[3]{1-x^2},\,b=\sqrt[3]{1-y^2}$. If $|a-b|\lt\varepsilon$, $|a^3-b^3+3ab(b-a)|\lt\varepsilon^3$, which is always true if $|a^3-b^3|+3|ab||b-a|\lt\varepsilon^3$, and always true if: $|y-x||y+x|+3\varepsilon|ab|\lt\varepsilon^3$ or: $$\frac{\delta}{\varepsilon}|y+x|+3|ab|\lt\varepsilon^2$$

Checking this region graphically, that region does account for the missing piece of the puzzle: finding suitable delta for when $|y|$ and $|x|$ are both the same side of $1$ is straightforward, and the tricky region where it is hardest to squeeze $|a-b|$ below $\varepsilon$ matches the set of points that satisfy the above inequality extremely well. Desmos does have numerical error of course, but it would seem that solving the above inequality give a $\delta$ as good as one can get! After some time, I was not able to push this inequality any further to find a closed form expression for $\delta$ that actually worked. What can I try? Any insights would be appreciated.
Below is a graph. The dark blue region is the set of all points satisfying $|a-b|\lt\varepsilon$, and that red region is the set of all points satisfying my reduced inequality. On close inspection they appear to match at the boundaries almost as perfectly as numerical precision will allow. The dark lines mark $y,x=1$, and in the bottom left, top right corners I have already solved the task; it is in the bottom right and top left where my inequality (as of yet unsolved) completes the picture.

Note: An empirical search finds $\delta=\varepsilon^3/8.00294102$ to be optimum (Desmos won't let me zoom in any deeper!)
 A: You could turn this into a calculus problem. Let $f(x) = \sqrt[3]{1-x^2}$ and $\varepsilon > 0$ be given. First, you have  $f''(x) < 0$ on the whole of $[-1,1]$ and so $g(x) = f'(x)$ is a decreasing function. You can also assume without loss of generality, $x < y$ and $y-x = \delta.$ Then, you can consider:
$$h(\delta) = \max\limits_{x\in[-1, 1-\delta]}|f(x) - f(x+\delta)|\quad  (1)$$
and then solve for:
$$h(\delta) = \varepsilon,$$
which would give some cubic equation in terms of $\delta.$ Note that $(1)$ needs to be divided in cases since $f$ is increasing on $[-1,0]$ and decreasing on $[0,1].$ Therefore you have three separate cases to consider:
$$\begin{cases}0\leq x<x+\delta\leq 1\\
-1\leq x<0\leq x+\delta\leq 1\\
-1\leq x<x+\delta\leq 0\end{cases}.$$
For example, I will do the first case here. Since $f$ is decreasing on $[0,1],$ we have $|f(x) - f(x+\delta)| = f(x)-f(x+\delta)$ and its derivative:
$$f'(x) - f'(x+\delta) > 0$$
since $f'(x)$ is also a decreasing function. Therefore, the maximum is attained at $x = 1-\delta:$
$$h(\delta) = f(1-\delta)-f(1) = \sqrt[3]{1-(1-\delta)^2}.$$
This would imply that for the first case, you get a your tightest possible $\delta$ as:
$$\delta(\varepsilon) = 1 - \sqrt{1-\varepsilon^3}.$$ Then, you can do the remaining two cases and find a single answer.
