Prove that there exists no bijective function $f: \Bbb{N} \to \Bbb{N}$ such that $f(mn)=f(m)+f(n)+3f(m)f(n)$ for $m,n \geqslant1.$ Prove that there exists no bijective function $f: \Bbb{N} \to \Bbb{N}$ such that $$f(mn)=f(m)+f(n)+3f(m)f(n)$$ for $m,n \geqslant1.$

This was a problem from a Putnam practice book and I couldn't seem to figure out how to show this. Initially, I started to wonder that if the function were bijective then it has to be injective as well as surjective. So trying to determine if its injective let $a,b \in \mathbb{N}$ then if it were injective $$f(mn)=f(ab) \implies f(m)+f(n)+3f(m)f(n) =f(a)+f(b)+3f(a)f(b)$$ but this doesn't seem to help since I don't have explicitly $f$. How should I look at this problem?
Let's assume $0\in\mathbb{N}$, since otherwise the condition $m,n\ge1$ would be redundant. Then, $m=n=1$ implies $f(1)=0$. If we define $g(n)=3\,f(n)+1$, we obtain $$g(mn)=g(m)g(n).\tag{mult}$$ So we have a completely multiplicative bijection between $\mathbb{N}$ and $\mathbb{M}=\{m\in\mathbb{N}: m=1\bmod3\}$. Now the numbers $10,34,55,187$ all belong to $\mathbb{M}$, so there must be natural numbers $n_1, n_2, n_3, n_4$ with $$g(n_1)=10,\quad g(n_2)=34,\quad g(n_3)=55,\quad g(n_4)=187.$$ The $n_i$ must be different, and they must be primes, because $d|n_i$ implies $g(d)|g(n_i)$ through (mult), and $g(n_i)$ by definition don't have non-trivial divisors in $\mathbb{M}$. But then, we have
$$g(n_1n_4)=g(n_1)\,g(n_4)=g(n_2)\,g(n_3)=g(n_2n_3),$$ and since $g$ is injective, $$n_1n_4=n_2n_3,$$ but this is impossible for four different primes.
1.Show that the function is well defined
2.Prove that there exist a kernel.
3. Prove that identity maps to element of the kernel
If you can't prove 1 and 2 the function or mapping is not bijective.
In other words, you can prove that the function is both injective and surjective( i.e if the inverse of the function maps back to every element in the domain) as required.