Let $f:H\to H' $ be a smooth map, between two Hilbert spaces. Suppose that $f$ is a local diffeomorphism, i.e. $f: B_\epsilon\to B'_{\epsilon'}$ is differentiable, and there exists $g:B'_{\epsilon'} \to B_\epsilon$, such that $f(g)=I$, and $$\|Df(x)^{-1}\|,\|Dg(y)\|\leq C_1,\|D^2g(y)\|\leq C_2.$$
We know that there exists a constant $c$ such that there exists a unique $h\in H$ $\|h\|_H\leq c$, we have $f(h)=0$.
Q: How to show that $$\|h\|_H\leq const \|f(0)\|_{H'}.$$
Being not knowledgeable enough to solve the problem as is, I had to additionally assume $g(f)(0) = 0$. This makes things considerably easier.
Since $h$ is unique, you know that $h = g(0)$. Hence $$ \|h\|_{H} = \|g(0)\|_{H} \leq \|g(0)- g(f(0))\|_{H} + \|g(f(0))\|_{H}. $$ You have by assumption that $g(f(0)) = 0$ and thus $$ \|h\|_{H} \leq \|g(0)- g(f(0))\|_{H}. $$ Finally, we have $\|g(0)- g(f(0))\|_{H} \leq C_2 \| f(0)\|_{H'}$ by the mean value theorem for Gateaux derivatives, which you can find everywhere.