" wrote: define the function $ G: 3\mathbb{N}_0 + 1\to 4\mathbb{N}_0 + 1$ as: $ g(x) = 4f\left(\frac {x - 1}3\right) + 1$ now note that $ f$ is bijective and so is $ g$. now let $ m,n\in\mathbb{N}_0$,we have: $ \begin{eqnarray*}g\left((3m + 1)(3n + 1)\right) & = & g\left(3(3mn + m + n) + 1\right) \\ & = & 4f\left(3mn + m + n\right) + 1 \\ & = & 4\left(4f(m)f(n) + f(m) + f(n)\right) + 1 \\ & = & \left(4f(m) + 1\right)\left(4f(n) + 1\right) \\ & = & g(3m + 1)g(3n + 1)$ so for every $ x,y\in 3\mathbb{N}_0 + 1$ we have: $ g(xy) = g(x)\cdot g(y)$ now notice that if we find a bijective function $ g: 3\mathbb{N}_0 + 1\to 4\mathbb{N}_0 + 1$ such that $ g(xy) = g(x)\cdot g(y)$,then the function $ f$ can be found as $ f(x) = \frac {g(3x + 1) - 1}4$. now let $ P_1,P_2$ be the sets of prime numbers of the form $ 3k + 1,3k + 2$ respectively and let $ Q_1,Q_2$ be the sets of prime numbers of the form $ 4k + 1,4k + 3$,now we define the bijective function $ h$ from $ P_1\cup P_2$ to $ Q_1\cup Q_2$ such that it pairs every element of $ P_1$ with one element of $ Q_1$,and the same thing for $ P_2,Q_2$. now we define $ g$ as: $ g(1) = 1$ and for every number $ n\in 3\mathbb{N}_0 + 1$ greater than $ 1$,let $ n = \prod p_i$ ( $ p_i$'s are prime numbers),then let: $ g(n) = \prod h(p_i)$ so obviously $ g(xy) = g(x)g(y)$,now that we have found $ g$,the function $ f$ can be easily obtained. QED

I found this problem to be quite interesting. Here's my solution: We use the following (well known) substitution- $$axy+x+y=\frac{1}{a}(a^2xy+ax+ay)=\frac{1}{a}((ax+1)(ay+1)-1)$$Then the given condition can be re-written as $$f \left(\frac{(3m+1)(3n+1)-1}{3} \right)=\frac{(4f(m)+1)(4f(n)+1)-1}{4} \quad \forall m,n \in \mathbb{N}_0$$This motivates us to consider the function $\psi:3 \mathbb{N}-2 \rightarrow 4 \mathbb{N}-3$ (where $\mathbb{N}$ represents the set of natural numbers) given by $$\psi(x)=4f \left(\frac{x-1}{3} \right)+1 \text{ and } \psi(mn)=\psi(m)\psi(n) \quad \forall m,n,x \in \mathbb{N}$$We are supposed to show that there exists such a bijective function $\psi$. Note that we have $\psi(1)=1$. Talking in group theoretic terms, we have to show that the commutative monoids $X=(3 \mathbb{N}-2, \times,1)$ and $Y=(4 \mathbb{N}-3, \times,1)$ are isomorphic. As we have to find a multiplicative function, we think about the canonical representation of the positive integers which are $1$ mod $3$. Let $\zeta$ be an involutive function sending the set of primes to itself. Then we consider the function given below. $$\text{For } n=\prod_{i=1}^{k} p_i^{\alpha_i} \text{, we have } \psi(n)=\prod_{i=1}^k (\zeta(p_i))^{\alpha_i}$$It's easy to see that this function is multiplicative. Also it is bijective (because $\zeta$ is bijective). So all we need to make sure is that the range of this function is $4 \mathbb{N}-3$. For this purpose, we include the following restriction on $\zeta$- $$\text{If } p \equiv a \pmod{3} \text{, then } \zeta(p) \equiv a \pmod{4} \text{ for } a \in \{-1,1\}$$Combining all this information, we get that $X$ and $Y$ are isomorphic, as desired. Hence, done. $\blacksquare$

Hopefully correct PEN solution.

Let $g:3\mathbb N_0+1\to4\mathbb N_0+1$ be defined by $g(3n+1)=4f(n)+1$ for all $n\in\mathbb N_0$, then the assertion implies that $g$ is multiplicative. Since $f$ is bijective, $g$ is also bijective. Let $h:\mathbb N\to\mathbb N$ be a multiplicative function defined on primes by: If $p=3k+1$ for $k\in\mathbb N_0$, then $h(3k+1)=4k+1$. If $p=3k+2$ for $k\in\mathbb N_0$, then $h(3k+2)=4k+3$. Restricting the domain of $h$ to $3\mathbb N_0+1$, the codomain is restricted to $4\mathbb N_0+1$, we have a suitable construction for $g$.