How to show the derivative by using the limit definition?

To use the Liebnitz rule, nicely explained by Vlad, we need

$$\tag 1 \int_\Omega \partial_k (v^\alpha \circ f(k,z))\,d \mu (z) <\infty.$$

Think about this for a bit: If we had $\Omega, f, v,$ such that the integrand is not bounded for some $k,$ we could fashion a probability measure $\mu$ on $\Omega$ for which $(1)$ fails. So boundedness is something we need to look for.

Let $\mathcal U$ be the collection of functions $u\in C^1(0,\infty)$ that are positive, increasing, and concave. Let $\mathcal U_b$ be the set of bounded functions $u\in \mathcal U.$

Lemma: Suppose $u\in \mathcal U.$ Then

$$u'(x) \le \frac{u(x)}{x}\,\,\text {for all } x>0.$$

If in addition $u\in \mathcal U_b,$ then

$$u'(x)x \le \|u\|_\infty.$$

I'll leave the proof of this to you for now. It's fairly simple, but ask questions if you like.

Let's turn to $(1).$ We are given $v\in \mathcal U_b.$ This implies $v^\alpha \in \mathcal U_b.$ Thus by the lemma, we have

$$ (v^\alpha)'(x) x \le \|v^\alpha\|_\infty.$$

We also have $f(k,z) \in \mathcal U$ as a function of $k$ for each fixed $z.$ Thus the lemma gives

$$f'(k,z) \le \frac{f(k,z)}{k}.$$

It follows that

$$\tag 2(v^\alpha)'(f(k,z)) f'(k,z) \le (v^\alpha)'(f(k,z)) \frac{f(k,z)}{k} \le \frac{\|v^\alpha\|_\infty}{k}.$$

But the left side of $(2)$ is exactly the integrand in $(1)$. Thus for $k \ge k_0$ we have the uniform bound $\|v^\alpha\|_\infty/k_0$ on the integrands. Of course the constant function equal to to this bound belongs to $L^1(\mu),$ since we're in a probablity space. This allows us to use Leibnitz on each interval $(k_0,\infty),$ and gives the desired result.

Perhaps I am missing something out, but it seems to me that you can obtain your final result using Leibniz's rule as shown below.

$$ \begin{align*} g(k) := \left( \int_{\Omega} \big[ v\left( f( k, z ) \right) \big]^\alpha \mu (\mathrm{d}z) \right)^{\frac{1}{\alpha}} =I^\frac{1}{\alpha}\left(k\right)\qquad (0<\alpha <1). \end{align*} $$ where $$ I(k)=\int_{\Omega} v^\alpha \left( f( k, z ) \right) \mu (\mathrm{d}z) $$

Thus, if we apply regular chain rule we get

$$ \frac{\partial }{\partial k}\left[g\left(k\right)\right] =\frac{\partial }{\partial k}\left[I^\frac{1}{\alpha}\left(k\right)\right] =\frac{1}{\alpha}I^{\frac{1}{\alpha}-1}\left(k\right) \frac{\partial }{\partial k}\left[I \left(k\right)\right] $$

Then, using Measure theory form of Leibniz rule we write

$$ \frac{\partial }{\partial k}\left[I \left(k\right)\right] = \frac{\partial }{\partial k}\left[\int_{\Omega} \big[ v\left( f( k, z ) \right) \big]^\alpha \mu \left(\mathrm{d}z\right)\right] = \int_{\Omega} \frac{\partial }{\partial k}\big[ v\left( f( k, z ) \right) \big]^\alpha \mu\left(\mathrm{d}z\right) = \int_{\Omega} \alpha v^{\alpha-1}\left( f( k, z ) \right) v'\left( f( k, z )\right) \frac{\partial }{\partial k}\big[ f( k, z ) \big] \mu (\mathrm{d}z) = \alpha\int_{\Omega} v^{\alpha-1}\left( f \right) v'\left( f \right) f_k\left( k, z \right) \mu \left(\mathrm{d}z\right) $$

Substituting the last expression into previous formula for $g'(k)$ we get

$$ g'\left(k\right) = \frac{1}{\alpha} \left(\int_\Omega \nu^\alpha\left(f\right)\mu\left(\mathrm{d}z\right)\right)^{\frac{1}{\alpha}-1} \cdot \alpha\int_{\Omega} v^{\alpha-1}\left( f \right) v'\left( f \right) f_k\left( k, z \right) \mu \left(\mathrm{d}z\right) $$ $$ \boxed{g'\left(k\right) = \left(\int_\Omega \nu^\alpha\left(f\right)\mu\left(\mathrm{d}z\right)\right)^{\frac{1}{\alpha}-1} \int_{\Omega} v^{\alpha-1}\left( f \right) v'\left( f \right) f_k\left( k, z \right) \mu \left(\mathrm{d}z\right)} $$

where $f=f\left(k,z\right)$ and $f_k\left(k,z\right)=\dfrac{\partial f\left(k,z\right)}{\partial k}$.

Justification for Using Leibniz's rule

As pointed out in comments by the OP, I have not provided justification for using Leibniz's rule in measure-theoretical form:

Let $X$ is an open subset of $ \mathbb{R} $, and $\Omega$ is a measure space. Suppose $ F\colon X\times \Omega \rightarrow \mathbb {R}$ satisfies the following conditions:

• $ F(x,\omega )$ is a Lebesgue-integrable function of $\omega$ for each $ x\in X$.

• For almost all $\omega \in \Omega$, the derivative $ F_{x} $ exists for all $ x\in X$.

• There is an integrable function $\displaystyle \theta \colon \Omega \rightarrow \mathbf {R}$ such that $$\left\lvert F_{x}(x,\omega )\right\rvert\leq \theta (\omega )$$ for all $ x\in X$ and almost every $\omega \in \Omega $.

Then, for all $x\in X$, $$ {\frac {d}{dx}}\int _{\Omega }F(x,\omega )\,d\omega =\int _{\Omega }F_{x}(x,\omega )\,d\omega . $$

Let us verify the last condition for the integrand

$$ F\left(k,z\right)=\left[ v\big( f( k, z ) \big) \right]^\alpha , \qquad 0<\alpha<1. $$

Observe that $F\left(k,z\right)$ is superposition of three concave increasing functions (treating $k\mapsto f\left(k,z\right)$ as function of single variable $k$)

$$ F\left(\cdot\right) = F_1\circ F_2\circ F_3 = F_3\left(F_2\left(F_1\left(\cdot\right)\right)\right), $$

where

$$ \begin{aligned} F_1\left(\tau\right) &= f\left(\tau,z\right), &&\text{where $z$ is fixed} \\ F_2\left(\tau\right) &= v\left(\tau\right), &&\\ F_3\left(\tau\right) &= \tau^{\alpha}, && 0<\alpha<1 \end{aligned} $$

with $\tau$ being dummy variable. As an exercise, I propose you to try to show explicitly that superposition of $F\left(\cdot\right) = F_1\circ F_2\circ F_3 = v^\alpha\left( f\left( k, z \right) \right)$ is convex in $k$.

Recall that concave function of single variable has monotonically decreasing derivative, which means that $F'\left(\tau\right)\leq F'\left(\tau_0\right)$ whenever $\tau\leq\tau_0$. For example, we can get bound

$$ \begin{aligned} F'\left(\tau\right)&<F'\left(0\right)& \text{since}& &f \colon \mathbb{R}_+ \times \Omega \to \mathbb{R}_+ \implies \tau>0 \end{aligned} $$

where $F'\left(0\right)$ does not depend on $\tau$ (and possibly depends on $\omega$).

Thus, choosing nonnegative constant $k_0$ we can bound derivative as $$ \frac{\partial}{\partial k} \Big[v^\alpha\big( f\left( k, z \right) \big)\Big] \leq \left.\frac{\partial}{\partial k} \Big[v^\alpha\big( f\left( k, z \right) \big)\Big]\right\rvert_{k_0} \qquad\text{ whenever } k\geq k_0. $$

Thus the third condition is satisfied.