Like the comments and the answer suggest one way to show this is by Dominated Convergence Theorem. Also we will make use of the following lemma from undergraduate level real analysis:
Lemma: Let $F:\mathbb{R}\to\mathbb{R}$ be a function and $x_0\in\mathbb{R}$. Then $\lim_{x\to x_0}f(x)=L \iff [\forall \{x_n\}_n \subseteq \mathbb{R}: x_n\to x_0\implies f(x_n)\to L]. (\ast)$
This lemma states that we can think of a continuous limiting process as a discrete limiting process, provided that we take into account any discretization of it. The utility of the lemma stems from the observation that if we have a discrete limiting process then we can define an appropriate sequence of functions.
Result: Let $f:[0,1]\times[0,1]\to\mathbb{R}$ be such that $\forall y\in[0,1]:f(\cdot,y):[0,1]\to\mathbb{R}$ is measurable. Suppose $\forall (x,y)\in[0,1]\times[0,1]:\dfrac{\partial f}{\partial y}(x,y)$ exists, and that there is an integrable $g:[0,1]\to\mathbb{R}:$
$\forall(x,y)\in [0,1]\times[0,1]: \left|\dfrac{\partial f}{\partial y}(x,y)\right|\leq g(x).$
Then
$\forall y\in[0,1]:\dfrac{d}{dy}\left[\int_{[0,1]} f(x,y)dx\right] =\int_{[0,1]} \dfrac{\partial f}{\partial y}(x,y) dx.$
Proof of the Result: Fix $y\in[0,1]$. Let $\{h_n\}_n\subseteq\mathbb{R}-0: h_n\downarrow0$ and set $\forall n: f_n:[0,1]\to\mathbb{R}, f_n(x):=\dfrac{f(x,y+h_n)-f(x,y)}{h_n}.$
Note that $\forall n: f_n$ is measurable since it is a linear combination of measurable functions. Since the partial derivative with respect to $y$ at any $(x,y)$ exists we have
$\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty} \dfrac{f(x,y+h_n)-f(x,y)}{h_n} \stackrel{(\ast)}{=} \lim_{h\to 0} \dfrac{f(x,y+h)-f(x,y)}{h} \stackrel{\tiny\mbox{def}}{=} \dfrac{\partial f}{\partial y} (x,y).$
Thus $f_n(\cdot)\to \dfrac{\partial f}{\partial y}(\cdot,y)$ pointwise on $[0,1]$.
Let $x\in[0,1]$. Since $f_n(x)\to\dfrac{\partial f}{\partial y}(x,y), \exists N,\forall n\geq N: \left|f_n(x)-\dfrac{\partial f}{\partial y}(x,y)\right|<1$. Hence $\forall n\geq N:$
\begin{align} |f_n(x)|-\left|\dfrac{\partial f}{\partial y}(x,y)\right|\leq \left|f_n(x)-\dfrac{\partial f}{\partial y}(x,y)\right|<1 \\ |f_n(x)|< \left|\dfrac{\partial f}{\partial y}(x,y)\right|+1 < g(x)+1 \in L^1([0,1]). \end{align}
Thus we may assume, by discarding the first $N-1$ terms of $\{f_n\}_n$ if necessary (i.e. by considering the sequence $\{f_n\}_{n\geq N}$) that $\{f_n\}_n$ is dominated by an integrable function. Then by Lebesgue's Dominated Convergence Theorem, $\dfrac{\partial f}{\partial y}(\cdot,y)$ is integrable on $[0,1]$ and
$\int_{[0,1]}f_n(x)dx\to\int_{[0,1]}\dfrac{\partial f}{\partial y}(x,y)dx.$
Since the sequence $\{h_n\}_n$ converging to $0$ was arbitrary we may apply the lemma above:
\begin{align} \lim_{n\to\infty}\int_{[0,1]}f_n(x)dx &= \lim_{n\to\infty}\int_{[0,1]}\dfrac{f(x,y+h_n)-f(x,y)}{h_n} dx \\ &=\lim_{n\to\infty}\dfrac{\int_{[0,1]}f(x,y+h_n)dx- \int_{[0,1]} f(x,y)dx}{h_n}\\ &\stackrel{(\ast)}{=} \lim_{h\to0}\dfrac{\int_{[0,1]}f(x,y+h)dx- \int_{[0,1]} f(x,y)dx}{h} =\dfrac{d}{dy} \int_{[0,1]}f(x,y)dx. \end{align}
Finally since $y$ is arbitrary the result holds for any $y$.
P.S.: For reference purposes this is Exercise 4.4.36 of Royden & Fitzpatrick's Real Analysis, 4e (pp. 90-91).