Prove that this language is NP-Hard
It seems that you're trying to prove that $\#\mathrm{3SAT}$ is in $\mathrm{NP}$. Since $\#\mathrm{3SAT}$ is $\mathrm{\#P}$-complete, and $\mathrm{\#P}$ seems to be harder than anything in the polynomial hierarchy, it's very unlikely that $\#\mathrm{3SAT}\in\mathrm{NP}$.

The error in your attempted proof is that your certificate doesn't have polynomial length. A satisfying assignment has length $\Theta(|x|)$ (the formula can't have more variables than its length, but it could be of the form $(x_1\lor x_2\lor x_3) \land (x_4\lor x_5\lor x_6) \land\dots$). The number&nbsp;$y$ could have any value between zero and $2^{|y|}-1$, so your certificate has length about $|x|2^{|y|}$, which is exponential in the input size.

However, the question asks you to prove that it's $\mathrm{NP}$-hard, not that it's $\mathrm{NP}$-complete. You just need to prove that there's a polynomial-time reduction from some $\mathrm{NP}$-complete problem to $\#\mathrm{3SAT}$.