A counterexample in one dimension: take $\Omega:=(0,1)$ and $f_n(x):=\frac{\sqrt 2}{x+\frac{1}{n}}$. Then $f''_n(x)-f_n^3(x)=0$ while $\| f''_n \| _{2,\Omega}=\|f^3_n\|_{2,\Omega}=O(n^{5/2}) \\ .$

A counterexample in one dimension: take $\Omega:=(0,1)$ and $f_n(x):=\frac{\sqrt 2}{x+\frac{1}{n}}$. Then $f''_n(x)-f_n^3(x)=0$ while $\| f''_n \| _{2,\Omega}=\|f^3_n\|_{2,\Omega}=O(n^{5/2}) \, .$