Remember that $\phi_t(x)$ is the solution at time $t$ to the system $\dot{y} = f(y)$ with initial condition $y(0) = x$.
The following skips a lot of details, but hopefully gives some idea of one approach to 'computing' the derivative you required.
To compute ${\partial \phi_t(x) \over \partial x}$, consider the system $L(z,x) = 0$, where $L(z,x)(t) = x+ \int_0^t f(z(\tau)) d \tau -z(t)$ and
$L : C^n[0,t] \times \mathbb{R}^n \to C^n[0,t]$.
Some work shows that $({ \partial L(z,x) \over \partial z}h) (s) =
\int_0^s { \partial f(z(\tau)) \over \partial y} h(\tau) d \tau-h(s)$ (with $ s \in [0,1]$), some more work shows that the map $D h = { \partial L(z,x) \over \partial z}h$ is continuous, and that the equation $D h = b$ is invertible (with $b \in C^n[0,t]$), and so the operator ${ \partial L(z,x) \over \partial z}$ is invertible.
It is straightforward to see that $({ \partial L(z,x) \over \partial x}\delta)(s) = \delta$ (with $ s \in [0,1]$), and applying the implicit function theorem (see Kantorovich & Akilov, "Functional analysis", for example) to $L(z,x) = 0$ gives the existence of some $\zeta$ such that
$L(\zeta(a), a) = 0$ for $a$ in some neighbourhood of $x$. Furthermore,
${ \partial L(z,x) \over \partial z} { \partial \zeta(x) \over \partial x} = -{ \partial L(z,x) \over \partial x}$, and so, for any $\delta \in \mathbb{R}^n$, we have
$({ \partial L(z,x) \over \partial z} ({ \partial \zeta(x) \over \partial x} \delta))(s) = -\delta$.
Unwinding this, and letting $\eta = { \partial \zeta(x) \over \partial x} \delta $, gives $\eta$ as the solution to the equation
$\int_0^s { \partial f(z(\tau)) \over \partial y} \eta(\tau) d \tau-\eta(s) = -\delta$, or in other words, $\eta$ solves $\dot{\eta}(s) = { \partial f(z(s)) \over \partial y} \eta(s)$, with $\eta(0) = \delta$.
Suppose $\Phi(t_1,t_2)$ is the state transition matrix for $\dot{\eta}(s) = { \partial f(z(s)) \over \partial y} \eta(s)$, then we have
${ \partial \phi_t(x) \over \partial x} \delta = \eta(t) = \Phi(t,0) \delta$, the solution of the 'linearised' differential equation at time $t$.
Best Answer
Substitute $f(x) = \sum_{k=0}^\infty a_k x^k$. Then \begin{equation} f'(x) = \sum_{k=0}^\infty k a_k x^{k-1} = \sum_{k=1}^\infty k a_k x^{k-1} = \sum_{m=0}^\infty (m+1) a_{m+1} x^m, \tag{1} \end{equation} and \begin{equation} 2 x f(x) = \sum_{k=0}^\infty 2a_k x^{k+1} = \sum_{n=1}^\infty 2 a_{n-1} x^n.\tag{2} \end{equation} Equating $(1)$ and $(2)$ term by term, you obtain a recurrence relation for the coefficients $a_n$. The initial condition $f(0) = 1$ yields $a_0 = 1$. You can now use the recursive relation to prove the statement.