show that $y'=f(y)$ has one solution only, $f(y)=-y\ln y$ 
let $f(y)=-y\ln y$ if $0<y<1$ and $f(y)=0$ otherwise
prove that $y'=f(y)$ has only one solution such that $y(0)=c$

I guess $c$ is a fixed arbitrary constant.
it's easy to see that $f$ is non negative, so the solution $y$ must be increasing.
despite this wolfram alpha says that $y(x)=e^{e^{-x+c}}$ is a solution to differential equation somewhere, but this is decreasing!
what am I missing?
how can I solve this problem?
I tried proving that $f$ satisfies the lipschitz condition, but I couldn't do it.

If $c$ is in $(0,1)$, $f(y(0))\ne0$ hence, at least in a neighborhood $(-t_0,t_0)$ of $0$, $y'/f(y)=1$. Since $y'/f(y)=(-\ln\ln y)'$, this yields $\ln\ln y(t)=-t+\ln\ln y(0)=-t+\ln\ln c$ on $(-t_0,t_0)$, that is, $y(t)=c^{\exp(-t)}$.
For every $c$ in $(0,1)$, $y_c:t\mapsto c^{\exp(-t)}$ is defined on the whole real line, hence the maximal solution with initial condition $y(0)=c$ is $y=y_c$, defined on the whole real line. Note that $t\mapsto\mathrm e^{-t}$ is decreasing and $c\lt1$ hence $y_c$ is increasing on the whole real line.
In the general case, assume that there exists a nonzero maximal solution $\bar y$ starting from $\bar y(0)=c$. Then, $f(\bar y(s))\ne0$ at least for some $s$ hence $\bar y(s)$ is in $(0,1)$. Thus, the function $y:t\mapsto\bar y(s+t)$ solves the differential equation $y'=f(y)$ with initial condition $y(0)=\bar y(s)$. The study above shows that $y(t)=y(0)^{\exp(-t)}$, that is, $\bar y(s+t)={\bar y(s)}^{\exp(-t)}$ for every $t$ in the real line. Hence $c=\bar y(0)={\bar y(s)}^{\exp(s)}$. Since $\bar y(s)$ is in $(0,1)$ and $\exp(s)\gt0$, $c$ is in $(0,1)$.
Finally, if some nonzero solution exists, then $c$ is in $(0,1)$.
