Uniqueness solutions of $dx/dt = f^2(x) + e^{-t}$. Someone can help me in the following problem? Is a question of Zhang.

Let $f(x)$ be continuous for $x \in \mathbb{R}$, show that $dx/dt = f^2(x) + e^{-t}$ has the property of uniqueness of solution. 

First, I'm not sure that $f^2(x)$ referred composition or power. I tried both ways and I failed. 
My idea was repeated the proof of Cauchy-Picard Theorem showing that
$$ F(x(t)) = x(0) + \int_{0}^{t}\left[ f^2(x(t)) - e^{-t} \right] dt$$
is a contraction. 
But I end up getting to the point where it seems I have to use that $f^2(x)$ satisfy a Lipschitz condition. But that does not seem true both in the case of the case of a composition as in case of the square power. For example, consider $f(x) = \sqrt{|x|}$, we have $f^2(x)$, seen as composition is $\sqrt[4]{|x|}$, which don't satisfy a Lipschitz condition. And consider $f(x) = \sqrt[4]{|x|}$, we have $f^2(x)$, seen as square power, is $\sqrt{|x|}$, which don't satisfy a Lipschitz condition.

Some progress:
We have 
$$x'(t) = f^2(x) + e^{-t} \Leftrightarrow x'(t) - e^{-t} = f^2(x)  \Leftrightarrow  (x(t) + e^{-t})'= f^2(x) $$
By the variable change $y(t) = x(t) + e^{-t}$, we have
$$y'(t) = f^2(y(t) - e^{-t}) = g^2(y(t))$$
See that the last step is valid both to composition case as in the power case. So, we have reduced the problem for the uniqueness of the problem
$$ y' = g^2(y), $$
where $g$ is merely a continuous function.
I do not know about you, but I think that this is not true. 
I think that $g(y) = \sqrt[4]{y}$, for $y\geq 0$ e $g(y)$, for $y<0$, is a counterexample in the power case.
 A: In the case that $f^2(x)$ means the square of $f(x)$, I have the following idea. Consider a local solution $x(t)$ of this differential equation on some interval $[a-r,a+r]$, $r>0$, satisfying some initial condition $x(a)=b$. The right hand side of the differential equation is positive and therefore $x$ is a strictly increasing function mapping  $[a-r,a+r]$ to some interval $[c,d]$, say, where $c<b<d$.
Consider its inverse function $z=x^{-1}$ defined on $[c,d]$ which is also differentiable. As $z(x(t))=t$
for all $t$ in $[a,b]$, differentiation gives $z'(x(t))x'(t)=1$. Replacing $t=z(s)$, $s$ in $[c,d]$, gives $z'(s)=1/x'(z(s))$. Now we use the differential equation $x'(t)=f^2(x(t))+e^{-t}$, replace $t=z(s)$ and find that $z$ satisfies the differential equation
$$z'(s)=\frac1{f^2(s)+\exp(-z(s))}$$
and the initial condition $z(b)=a$. The above differential equation satisfies the hypothesis of Picard-Lindelöf's theorem, because the right hand side has a continuous partial derivative with respect to $z$.
Therefore it has a unique solution, not only locally but also globally.
Hence the function $x(t)$ we considered in the beginning must be the
inverse of the unique solution of the $z$-differential equation satisfying $z(b)=a$. This determines $x=x(t)$ uniquely.
To complete this approach, we can show that the solution of the $z$-differential equation (with initial condition) is stricly increasing and has therefore an inverse function.
EDIT: The right hand side of the $z$-differential equation is $F(s,z)=1/(f^2(s)+\exp(-z))$ and continuous on $R^2$ with a continuous partial derivative with respect to $z$. By a theorem based on Picard-Lindelöf, for every initial condition $z(b)=a$, there is a unique solution $z:]c,d[\to R$ having maximal interval of definition ($c,d$ can be infinite, $c<b<d$). Either $c=\infty$ or $\lim_{s\to c} |z(s)|=\infty$ or both.
As $z'(s)$ is positive, $z$ is strictly increasing and hence its image $z(]c,d[)$ is also an open interval, say $]A,B[$. Its inverse function $x:]A,B[\to R$ is then also differentiable and, by a calculation analogous to the above, is a solution of $x'=f^2(x)+e^{-t}$, $x(a)=b$. Actually it is the unique solution having a maximal interval of definition. 
