Your solution is not correct. You have to show that $\mathbf{\text{negl}_2}$ satisfies the definition of negligible functions and what you "showed" actually is that given any sufficiently large polynomial $p'(n)$, it holds that $\mathbf{\text{negl}_2} < \frac{p(n)}{p'(n)}$.

There are to easy ways of solving that exercise:

• Try to prove it by contradiction supposing that $\mathbf{\text{negl}_2}$ is not negligible and then finding a polynomial whose inverse is asymptotically smaller than $\mathbf{\text{negl}_1}$.
• Try to prove it directly: from any given polynomial $p'(n)$, you know that $p(n)p'(n)$ is a polynomial, therefore, $\mathbf{\text{negl}_1}$ must be smaller than $1 / (p(n)p'(n))$ for all $n$ bigger than some $n_0$...

Your solution is not correct. You have to show that $\mathbf{\text{negl}_2}$ satisfies the definition of negligible functions and what you "showed" actually is that given any sufficiently large polynomial $p'(n)$, it holds that $\mathbf{\text{negl}_2} < \frac{p(n)}{p'(n)}$.

I see two easy ways of solving that exercise:

• Try to prove it by contradiction supposing that $\mathbf{\text{negl}_2}$ is not negligible and then finding a polynomial whose inverse is asymptotically smaller than $\mathbf{\text{negl}_1}$.
• Try to prove it directly: from any given polynomial $p'(n)$, you know that $p(n)p'(n)$ is a polynomial, therefore, $\mathbf{\text{negl}_1}$ must be smaller than $1 / (p(n)p'(n))$ for all $n$ bigger than some $n_0$...