Let $\phi(t, x_0)$ be a solution of the one-dimensional differential equation $$\dot{x}= f(x),$$ with $\phi(0, x_0) = x_0$. Show that its derivative $\frac{\partial}{\partial x_0}\phi(t, x_0)$ is given by $$\frac{\partial}{\partial x_0}\phi(t, x_0) = exp \Big(\int_{0}^{t}f'(\phi(s, x_0))ds\Big)$$
Here in my proposed solution.
By the fundamental theorem of calculus, we have
$$\phi(t, x_0)=x_0 + \int_{0}^{t}f(\phi(s, x_0))ds$$
because $\frac{\partial{\phi}}{\partial t}(t, x_0)=f(\phi(t, x_0))$ and $\phi(0, x_0)=x_0$.
If we differentiate this solution with respect to $x_0$, we obtain via the chain rule that
$$\frac{\partial{\phi}}{\partial x_0}(t, x_0)=1 + \int_{0}^{t}\frac{\partial{f}}{\partial{x_0}}(\phi(s, x_0)) \cdot \frac{\partial{\phi}}{\partial x_0}(s, x_0)ds$$
Let $$z(t)= \frac{\partial{\phi}}{\partial x_0}(t, x_0)$$
Then,
$$z(0)= \frac{\partial{\phi}}{\partial x_0}(0, x_0)=1$$
by the analysis above. Therefore, if we differentiate $z$ with respect to $t$, we find that
\begin{equation} \begin{split} z'(t) & = \frac{\partial{f}}{\partial{x_0}}(\phi(t, x_0)) \cdot \frac{\partial{\phi}}{\partial x_0}(t, x_0)\\ & = \frac{\partial{f}}{\partial{x_0}}(\phi(t, x_0)) \cdot z(t) \end{split} \end{equation}
We do not have an explicit solution to $\phi(t,x_0)$, but the above equation tells us that $z(t)$ solves the following differential equation,
$$z'(t)= \frac{\partial{f}}{\partial{x_0}}(\phi(t, x_0)) \cdot z(t)$$
Therefore,
$$\frac{dz}{dt}= \frac{\partial{f}}{\partial{x_0}}(\phi(t, x_0)) \cdot z(t)$$
Rearranging terms produces
$$\frac{1}{z(t)}{dz}= \frac{\partial{f}}{\partial{x_0}}(\phi(t, x_0))dt$$
Hence,
$$ln(z(t))= \int_{0}^{t}\frac{\partial{f}}{\partial{x_0}}(\phi(s, x_0))ds$$
And taking the exponential of both sides produces
$$z(t)= exp \Big(\int_{0}^{t}\frac{\partial{f}}{\partial{x_0}}(\phi(s, x_0))ds\Big)$$
As $z(t)=\frac{\partial{\phi}}{\partial x_0}(t, x_0)=exp \Big(\int_{0}^{t}f'(\phi(s, x_0))ds\Big)$, we are done.
I'm not certain if there is a more direct approach. Please let me know if the solution can be improved.
From the comments above, here is a shorter answer:
As $\dot{x}=f(x) \implies \frac{dx}{dt}=f(x)$, we can differentiate both sides with respect to $x_0$ to form
$$\frac{d}{dt}\frac{\partial{x}}{\partial{x_0}}=\frac{df(x)}{dx}\frac{\partial{x}}{\partial{x_0}}=f'(x)\frac{\partial{x}}{\partial{x_0}}$$
Therefore, as $\phi(t,x_0)$ is a solution to the ODE, we know that $x(t)=\phi(t,x_0)$. Hence,
$$\frac{d}{dt}\frac{\partial\phi}{\partial{x_0}}(t,x_0)=f'(\phi(t,x_0))\frac{\partial\phi}{\partial{x_0}}(t,x_0)$$
Now, let $z(t)=\frac{\partial\phi}{\partial{x_0}}(t,x_0)$. Then,
$$\frac{d}{dt}z(t)=f'(\phi(t,x_0))z(t)$$
So,
$$\frac{1}{z(t)}d{z(t)}=f'(\phi(t,x_0))dt$$
Hence, by the fundamental theorem of calculus,
$$ln(z(t))=\int_0^t{f'(\phi(s,x_0))ds}$$
Therefore, if we take the exponential of both sides,
$$z(t)=exp\Big(\int_0^t{f'(\phi(s,x_0))ds}\Big)$$
So, $\frac{\partial\phi}{\partial{x_0}}(t,x_0)=exp\Big(\int_0^t{f'(\phi(s,x_0))ds}\Big)$.