Let $f: E \rightarrow \mathbb{R}^n$ be a locally Lipschitz vector field for Initial Value Problem (IVP) $\dot{y} = f(y)$ with $y(0)=x$, where $E \subset \mathbb{R}^n$ is open.
Definition: Let $\phi : \Omega \rightarrow E $ be the solution to the IVP $\dot{y} = f(y)$ with $y(0)=x$ where $\Omega=\{(t,x) \mid x \in E , t \in I_x\}$ and $I \in \mathbb{R}$ is an interval which includes $0$.
Prove flow of vector field $(-f(y))$ is $\phi_{-t}(x)$ where $\phi_{t}(x)$ is the flow of $(f(y))$?
My try:
When $u(t)$ is the solution to the IVP, it satisfies the following:
$$ u(t)=x+\int_0^t f(u(s))ds $$
Now let $\phi_t(x)$ be the solution so
$$ \phi_t(x)=x+\int_0^t f(\phi_s(x))ds \tag{1} $$
Now suppose we have $\dot{y} = -f(y)$ with $y(0)=x$ as our IVP. Let $\phi'_t(x)$ be the solution to the new IVP so
$$ \phi'_t(x)=x+\int_0^t -f(\phi'_s(x))ds $$
which is equivalent to
$$ \phi'_t(x)=x+\int_t^0 f(\phi'_s(x))ds \tag{3} $$
How would you show $\phi'_t(x)=\phi_{-t}(x)$ from $(3)$ and $(1)$.