(This answer doesn't address the question for the whole curve, but really just the question of solving the differential equation near $0$).

I think one way to proceed is as follows, which can be made rigorous and is how things work when $n = 2:$ let $f(t) = \int_{0}^{t} \frac{1}{(1- u^{n})^{1/n}}du$ for $ t \in (-1,1),$ where we take the unique real positive $n$-th root, and let $g$ be the inverse function to $f,$ which exists as $f$ is increasing where defined. Note that $f(0) = 0$ and that $f$ is an odd function if $n$ is even. Note also that (by the Fundamental Theorem of Calculus) $f$ is infinitely differentiable on $(-1,1).$

Then $f(g(t)) = t.$ Now $g^{\prime}(t) = \frac{1}{f^{\prime}(g(t))}$ and
$f^{\prime}(g(t)) = \frac{1}{(1-g(t)^{n})^{\frac{1}{n}}}.$ Hence we do have
$g(t)^{n} + g^{\prime}(t)^{n} = 1.$ Note also that $g(0) = 0$ and $g^{\prime}(0) = 1.$ From this, and the equation $g(t)^{n} + g^{\prime}(t)^{n} = 1,$ we can inductively determine the Maclaurin series for $g$.

Later edit partially motivated by Christian Remling's comment. The only difficulty seems to be to find ( near enough $0$) a good choice of $f$ in the complex case, and show that it is injective in neighbourhood of $0$: For the sake of completeness, I outline a proof that in the complex case, we can construct a solution $g(z)$ with $g(z)^{n} + g^{\prime}(z)^{n} =1,$ and $g(0) = 0,g^{\prime}(0) = 1,$ and that $g(z)$ has a Maclaurin series with strictly positive radius of convergence.

We define $f$ on the open unit disk $C(0;1)$ by $f(z) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{1+nk}z^{1+nk} \left( \begin{array}{clcr} \frac{-1}{n}\\ k \end{array} \right).$ The power series does converge on $C(0;1),$ and we have
$f^{\prime}(z) = (1-z^{n})^{\frac{-1}{n}} = \sum_{k=0}^{\infty} (-1)^{k}z^{nk} \left( \begin{array}{clcr} \frac{-1}{n}\\ k \end{array} \right)$ for $z \in C(0;1).$

Let $r$ be a real number with $ 0 < r < (1-\frac{1}{n+1})^{\frac{1}{n}}.$
Then $0 < \frac{r^{n}}{n(1-r^{n})} < 1.$ It is easy to check from the power series that for $z,w \in C(0;r),$ we have

$|z-w| ( 1 - \frac{r^{n}}{n(1-r^{n})}) \leq |f(z)-f(w)| \leq |z-w| ( 1 + \frac{r^{n}}{n(1-r^{n})}).$ In particular, $f$ is injective on $C(0;r),$ so the inverse function, $g$ say, is defined on the set $f(C(0;r)),$ which is an open set by the Open Mapping Theorem. The above inequalities imply that $g$ is uniformly continuous on $f(C(0;r)),$ and then it follows that $g^{\prime}(z) = \frac{1}{f^{\prime}(g(z))} = (1-g(z)^{n})^{\frac{1}{n}}.$ Hence we do have $g(z)^{n} + g^{\prime}(z)^{n} = 1$ for $z \in f(C(0;r)).$ Since $g$ is differentiable on an open set which includes $0,$ its Maclaurin series has a strictly positive radius of convergence.

We also note that for any complex $n$-th root of unity, we have $f(\omega z ) = \omega f(z)$ for any $z \in C(0;r)$ and for any (not necessarily primitive) $n$-th root of unity $\omega,$ so that we also have $g(\omega z) = \omega g(z)$ for any such $\omega$ and any $z \in f(C(0;r)).$ It follows easily from this that the Maclaurin series for $g(z)$ has the form $z + \sum_{k=1}^{\infty} a_{k} z^{1+nk}.$