Let $f$ be a function $f : \mathbb{N} \to \mathbb{N}$ such that $f(2x+3y)=f(x)f(y)$, determine $f$ here what I did .
$$f(0)=f(0)^2$$
so $f(0)=1$ or $f(0)=0$
IF $f(0)=1$
we have $f(2y)=f(y)$
$$f(1)=f(2)=\ldots=f(2^n)=a$$
the equation $f(x)-a=0$ has infinitly many solutions , so $f(x)=a$
since f(0)=1 , $f(x)=1$.
i don't know how to handle the other case , i just found that $f(2x)=f(3x)=0$

This is related to the Frobenius problem (or coins or Mcnuggets problem) of two numbers:
If $a,b\in\mathbb{N}$ are two numbers such that $\gcd(a,b)=1$, then all numbers $n\geq(a-1)(b-1)$ are representable as a nonnegative combination of $a,b$, i.e., $n=ax+by$ with $x,y\in\mathbb{N}$.
1) If $f(0)=0$: 
Since $\gcd(4,3)=1$ and $(4-1)(3-1)=2$, if $n\geq6$, then there are $x_n,y_n\in\mathbb{N}$ such that $n=4x_n+3y_n$.
Since $f(0)=0$ we get $f(2x)=f(2x+3\cdot 0)=f(x)f(0)=0$ for every $x\in\mathbb{N}$, so $f(n)=f(4x_n+3y_n)=f(2x_n)f(y_n)=0$ for all $n\geq6$.
Since $f(0)=f(2)=f(4)=0$, it just remains to show what happens with $f(1),f(3),f(5)$.
Note that $f(3)^2=f(3)f(3)=f(2\cdot 3+3\cdot 3)=f(15)=0$ (because $15\geq6$), so $f(3)=0$. Similarly, $f(5)^2=f(25)=0$, and finally $f(1)^2=f(5)=0$, so $f(1)=f(3)=f(5)=0$ and $f(x)=0$ for every $x\in\mathbb{N}$.
b) If $f(0)=1$:
Since $\gcd(2,3)=1$, for every $n\geq2$ there are $x_n,y_n\in\mathbb{N}$ such that $n=2x_n+3y_n$. Denote $a:=f(1)$. By strong induction, suppose that for every number $m<n$, $f(m)$ is some power of $a$ (we have as base cases $f(0)=a^0$, $f(1)=a^1$). Then $f(n)=f(2x_n+3y_n)$ with $0\leq x_n,y_n<n$, thus $f(n)=f(x_n)f(y_n)=a^ra^s=a^{r+s}$ for some exponents $r,s$.Note that we can do this in such a way that the exponent is never $0$ (except for $f(0)$).
On the other hand, the first number representable in two ways (with different numbers unregarding the order) as a nonnegative linear combination of $2$ and $3$ is $8=2\cdot 4+3\cdot 0=2\cdot 1+3\cdot 2$; thus $f(8)=f(4)f(0)=f(4)$ and $f(8)=f(1)f(2)$, so $f(4)=f(1)f(2)=af(2)$. But $f(4)=f(2\cdot 2 +3\cdot 0)=f(2)f(0)=f(2)$, hence $af(2)=f(2)$, so either $a=1$ or $f(2)=0$; and $f(2)=f(2\cdot 1+3\cdot 0)=f(1)=a$, so either $a=1$ or $a=0$. Noting that every other image is a power of $a$, we get either $f(x)=1$ for every $x$ or $f(0)=1$, $f(x)=0$ for every $x>0$.
Observe that these functions indeed satisfy the given condition.
