Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$? Suppose that we are given $\mathbb{Z}_{N}$ and an element $x^u \in \mathbb{Z}_{N}$ with $u \in (0,l]$ where $l$ is the bit-size of $N$. Is it difficult to recover $x$ by knowing $u$ without knowing the factorization of $N$?
Yes, the problem of finding unknown random $x\in\mathbb Z_N$ given $N\in\mathbb N$, $u\in\mathbb N$ with $1&lt;u\le\lceil\log_2N\rceil$, and $x^u$ computed in $\mathbb Z_N$, is believed hard unless the factorization of $N$ can be determined, which itself is believed hard for appropriately constructed RSA moduli $N$. Moreover, depending on $u$ and $N$, there might be several solutions with no way to tell the right one (if $u$ is odd and $N$ squarefree with no divisor up to $\lceil\log_2N\rceil$ then there is a single solution). Notice that I have excluded $u=1$ (which the question allows), for it makes finding $x$ trivial.

When $u$ is odd this is the RSA problem with small exponent, which is believed hard when the factorization of $N$ can't be determined (with the caveat that a usual definition of that problem asks that $N$ is squarefree and without small divisors).

When $u$ is $2$, this is the square root modulo unknown composite problem, which is demonstrably hard when the factorization of $N$ can't be determined.

Ability to consistently solve both that square root problem and the RSA problem implies ability to find a solution to the question's problem, by writing $u$ as $e\cdot2^s$ with odd $e$, solving the RSA problem for exponent $e$, and solving the square root problem $s$ times.



Addition: this answer is for version 5 of the question. As correctly noted in another answer for an earlier version of the question, knowing that $N$ is a square does not make the problem easier if the factorization of $N$ remains unknown.



Further addition following comments:


It is easy to find the inverse of any given element of $\mathbb Z_N$ when its exists, using the Extended Euclidian algorithm.
It is hard to exhibit $a\in\mathbb Z_N$ that has no inverse modulo $N$, other than $0$ or a multiple of an easily found factor of $N$ (argument: if such an $a$ was found, then $\gcd(a,N)$ would be a non-trivial divisor of $N$  contributing towards finding the full factorization of $N$).
If $\gcd(u,N)=1$ (which we'll assume hereafter), it is thus easy to find non-negative integers $v$ and $k$ with $u\cdot v=k\cdot N+1$, otherwise said with $v\equiv u^{-1}\pmod N$.
It does follow that $\forall x\in\mathbb Z_N$, $(x^u)^v=x^{k\cdot N+1}=(x^N)^k\cdot x$.
That does not seem to help finding $x$, because we have nothing telling us that $x^N=1$; on the contrary, $x^N=1$ seldom holds when $x\ne 1$; in particular $x^N=1$ holds only for $x=1$ when $N$ is prime, since in this case: $\forall x\in\mathbb Z_N$, $x^N=x$ by Fermat's little theorem.