Find the derivative of $\sqrt[n]{x}$ using the formal definition of a derivative

The question is definitely not trivial. +1 for OP. The solution follows from the following theorem:

Theorem: If $a > 0$ and $n$ is a rational number then $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ This is one of the standard limits which can be used to evaluate many limits involving algebraic functions.

The proof of the above theorem is easy if $n$ is an integer. For positive integers we can simply use $$x^{n} - a^{n} = (x - a)\sum_{i = 0}^{n - 1}x^{n - 1 - i}a^{i}$$ For $n = 0$ the result is obvious. For negative integer $n = -m$ we can use $x^{n} = 1/x^{m}$ and the fact that the result holds for positive integers. Similarly if the result holds positive rational number $n$ we can show that it holds for negative rational $n$ also.

Thus we need to show that if $n = p/q$ with integers $p > 0, q > 1$ then the formula $(1)$ holds. Let $b = a^{1/q}$ so that $a = b^{q}$. We know that $$\lim_{y \to b}\frac{y^{q} - b^{q}}{y - b} = qb^{q - 1}\tag{2}$$ From $(2)$ it follows that the ratio $(y^{q} - b^{q})/(y - b)$ is bounded and away from $0$ as $y \to b$. Hence its reciprocal is also bounded and away from $0$ as $y \to b$. Also note that when $y \to b$ then $x = y^{q} \to b^{q} = a$ (and vice-versa because $f(y) = y^{q}$ is strictly monotone in $[0, \infty)$). Thus the ratio $(x^{1/q} - a^{1/q})/(x - a)$ is bounded when $x = y^{q} \to a$ and therefore $$\lim_{x \to a}x^{1/q} = a^{1/q}\tag{3}$$ (this by the way proves continuity of $x^{1/q}$).

Now we have \begin{align} L &= \lim_{x \to a}\frac{x^{n} - a^{n}}{x - a}\notag\\ &= \lim_{x \to a}\frac{x^{p/q} - a^{p/q}}{x - a}\notag\\ &= \lim_{t \to b}\frac{t^{p} - b^{p}}{t^{q} - b^{q}}\text{ (putting }x = t^{q}, a = b^{q}\text{ and using (3))}\notag\\ &= \lim_{t \to b}\dfrac{\dfrac{t^{p} - b^{p}}{t - b}}{\dfrac{t^{q} - b^{q}}{t - b}}\notag\\ &= \frac{pb^{p - 1}}{qb^{q - 1}}\notag\\ &= \frac{p}{q}b^{p - q}\notag\\ &= na^{n - 1}\notag \end{align} There is another way to prove this (via inequalities and squeeze theorem) without using the continuity of $x^{1/q}$. Let me know if you are interested in that version.

Update: On request of OP I am providing a proof of formula $(1)$ based on Squeeze Theorem. The credit for this proof must go to G. H. Hardy!

In what follows all the numbers are positive (whether they are integers, rationals or reals will be mentioned as and when needed).

Let $a, b$ be real numbers with $a > 1 > b > 0$. Let $r$ be an integer. Clearly we have $a^{r} > a^{i}$ for all $i = 0, 1, 2, \ldots, r - 1$. Hence on adding these inequalities we get $$ra^{r} > 1 + a + a^{2} + \cdots + a^{r - 1}$$ Multiplying by $(a - 1) > 0$ we get $$ra^{r}(a - 1) > a^{r} - 1$$ Adding $r(a^{r} - 1)$ on both sides, and dividing by $r(r + 1)$, we obtain $$\frac{a^{r + 1} - 1}{r + 1} > \frac{a^{r} - 1}{r}\tag{4}$$ Similarly we can prove that $$\frac{1 - b^{r + 1}}{r + 1} < \frac{1 - b^{r}}{r}\tag{5}$$ It follows that if $r, s$ are positive integers with $r > s$ then $$\frac{a^{r} - 1}{r} > \frac{a^{s} - 1}{s},\,\frac{1 - b^{r}}{r} < \frac{1 - b^{s}}{s}\tag{6}$$ If we put $s = 1$ we get $$a^{r} - 1 > r(a - 1),\, 1 - b^{r} < r(1 - b)\tag{7}$$ for $r > 1$.

Next we show that the inequalities $(6), (7)$ hold when $r, s$ are positive rational numbers with $r > s$. Let $r = k/l, s = m/n$ and $r > s$ implies that $kn > lm$. Let $c = a^{1/ln}$ so that $c > 1$. In the first inequality of $(6)$ we can replace $a$ by $c$, $r$ by $kn$ and $s$ by $lm$ to get $$\frac{c^{kn} - 1}{kn} > \frac{c^{lm} - 1}{lm}$$ or $$\frac{a^{r} - 1}{r} > \frac{a^{s} - 1}{s}$$ In similar manner we can prove that other inequalities also hold when $r, s$ are rational numbers. Now that $r, s$ are rational, it is possible to take $r = 1$ in $(6)$ to get $$a^{s} - 1 < s(a - 1),\,1 - b^{s} > s(1 - b)\tag{8}$$ for rational $s$ with $0 < s < 1$. Thus we have inequalities $(6)-(8)$ for all positive rational numbers $r, s$ with $r > 1 > s$.

In what follows we will assume that $a, b$ are real with $a > 1 > b > 0$ (same as before) and $r, s$ are rational with $r > 1 > s > 0$. Clearly $1/b > 1$ and hence replacing $a$ by $1/b$ and $b$ by $1/a$ in $(7)$ we get $$a^{r} - 1 < ra^{r - 1}(a - 1),\, 1 - b^{r} > rb^{r - 1}(1 - b)\tag{9}$$ Similarly from $(8)$ we get $$a^{s} - 1 > sa^{s - 1}(a - 1),\, 1 - b^{s} < sb^{s - 1}(1 - b)\tag{10}$$ Combining $(7)$ and $(9)$ we get $$ra^{r - 1}(a - 1) > a^{r} - 1 > r(a - 1)\tag{11}$$ Writing $a = x/y$ we get $$rx^{r - 1}(x - y) > x^{r} - y^{r} > ry^{r - 1}(x - y)\tag{12}$$ for $x > y > 0$. Similarly from $(8)$ and $(10)$ we get $$sx^{s - 1}(x - y) < x^{s} - y^{s} < sy^{s - 1}(x - y)\tag{13}$$ for $x > y > 0$.

From the above inequalities it is clear that the function $f(x) = x^{r}$ is continuous for $x > 0$. Taking reciprocals it is easy to see that the function $f(x)$ is continuous even if $r$ is negative rational number. Further if we divide by $(x - y) > 0$ and let $x \to y^{+}$ we get via Squeeze Theorem the fundamental result $$\lim_{x \to y^{+}}\frac{x^{r} - y^{r}}{x - y} = ry^{r - 1}$$ for all positive rational numbers $r$ and $y > 0$. Interchanging the roles of $x, y$ it is easy to see that the limit holds for $x \to y^{-}$. This proves the formula $(1)$ for positive rational values of $n$.

This is the way Hardy proves the formula $$\frac{d}{dx}(x^{n}) = nx^{n - 1}$$ for rational $n$ in his classic text "A Course of Pure Mathematics".

Using the identity

$$ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \dots + ab^{n-2} + b^{n-1}) $$

you have

$$ \frac{(x+h)^{\frac{1}{n}} - x^{\frac{1}{n}}}{h} \cdot \frac{(x+h)^{\frac{n-1}{n}} + (x+h)^{\frac{n-2}{n}}x+\dots+(x+h)^{\frac{1}{n}}x^{\frac{n-2}{n}}+x^{\frac{n-1}{n}}}{(x+h)^{\frac{n-1}{n}} + (x+h)^{\frac{n-2}{n}}x+\dots+(x+h)^{\frac{1}{n}}x^{\frac{n-2}{n}}+x^{\frac{n-1}{n}}} = \frac{1}{(x+h)^{\frac{n-1}{n}} + (x+h)^{\frac{n-2}{n}}x+\dots+(x+h)^{\frac{1}{n}}x^{\frac{n-2}{n}}+x^{\frac{n-1}{n}}} \xrightarrow[h \to 0]{} \frac{1}{x^{1-\frac{1}{n}}+x^{1-\frac{2}{n}}x^{\frac{1}{n}}+\dots +x^{\frac{1}{n}}x^{1-\frac{2}{n}}+x^{1-\frac{1}{n}}}=\frac{1}{n x^{1 - \frac{1}{n}}}=\frac{x^{\frac{1}{n}-1}}{n}. $$