show that in the one dimensional case, we have
$\dfrac{\partial \phi}{\partial x}(t,x) = \exp\displaystyle{\left(\int_{t_0}^{t}\dfrac{\partial f}{\partial x}(s,\phi(s,x))ds\right)}$.
I tried to solve using the IVP $x'=f(t,x).x \,\,\,\, , x(t_0)=x_0$ and I used $\phi$ is a solution. then replace the solution in the IVP and obtain $\dfrac{\phi '}{\phi}= f(t,\phi)$. From here, I did the integration of $t_0$ to $t$, then partially derived with respect to $x$. In the end I got this relationship
$\dfrac{\partial \phi}{\partial x} = \exp\displaystyle{\left(\int_{t_0}^{t}\dfrac{\partial f}{\partial x}(s,\phi(s))ds\right)}$.
I'm finding this question very strange because $\phi$ should not be of the form $\phi (t)$? here we have $\phi(t,x)$ in $\dfrac{\partial \phi}{\partial x}(t,x) = \exp\displaystyle{\left(\int_{t_0}^{t}\dfrac{\partial f}{\partial x}(s,\phi(s,x))ds\right)}$. Can someone help?
If you are content with purely formal calculations, you can proceed as follows. We have $$ \frac{\partial \phi}{\partial t}(t, x) = f(t, \phi(t, x)), $$ which, after differentiating in $x$ and changing order of partials, gives $$ \frac{\partial}{\partial t} \Bigl( \frac{\partial \phi}{\partial x} \Bigr) (t,x) = \frac{\partial f}{\partial x}(t, \phi(t,x))\, \frac{\partial \phi}{\partial x}(t, x). $$ So, $$ \frac{\partial \phi}{\partial x} (t,x) $$ (with $x$ for the moment fixed) satisfies a linear ordinary differential equation $$ y'(t) = \frac{\partial f}{\partial x}(t, \phi(t,x))\, y(t). $$ What about the initial value? One has $\phi(0, x) = x$, so differentiating it in $x$ we get $$ y(0) = 1. $$ Therefore $$ \dfrac{\partial \phi}{\partial x}(t,x) = \exp\displaystyle{\left(\int_{t_0}^{t}\dfrac{\partial f}{\partial x}(s,\phi(s,x))\,ds\right)}. $$
Indeed, the above calculations are legitimate, since (assuming that $\partial f/\partial x$ is continuous) $\phi(\cdot, \cdot)$ is so regular that one can change the order of partial differentiations, etc. But that's a different story.