Given that:
$$ \begin{cases} -\Delta u(x)&=ku^2(x)(1-u(x)),&\text{ if }x \in \Omega \\ \phantom{-}u(x)&=0,&\text{ if }x \in \partial\Omega \end{cases} $$
and $k>0$ constant, show that $0 \leq u(x)\leq 1$.
I assumed that $\exists x_0 \in \Omega$ such as $u(x_0)>1$.
Using the conditions of maximum points of a function, by the Jacobian and Hessian of $u$, I showed that
$u_{xx}<0$ and $u_{yy}<0$ but a I couldn't find something about $u_{zz}$.
If I find that $u_{zz}<0$ then $-\Delta u(x_0)>0$ but $ku^2(x_0)(1-u(x_0))<0$ contradiction.
So $0\leq u(x)\leq 1$.
Note that if $x_0$ is a maximum point, then the function of a real variable $t \mapsto u(x_0 + t e_i)$ has a maximum at $t = 0$, so you shall have $\partial_{ii} u < 0$. No need for arguments with the Hessian matrix.