1. Find the set $A$

We need $x^2-10x+9\le 0.$ Factorizing, $x^2-10x+9=(x-1)(x-9).$ So, $(x-1)(x-9)\le 0 \implies 1\le x\le 9.$ Since $x\in \mathbb N$, $A=\{1,2,3,4,5,6,7,8,9\}.$ Thus, $|A|=9$.

2. Understand the codomain $B$

$B=\{n^2:n\in\mathbb N\}$ which is the set of natural-number squares.

For each $x\in A$, the value $f(x)$ must satisfy: $f(x)\in B \quad \text{and} \quad f(x)\le (x-3)^2+1.$ So for each fixed $x$, we count how many perfect squares are $\le (x-3)^2+1$.

3. Count choices for each $x$

Let $U(x)=(x-3)^2+1.$ We compute for each $x\in A$:

• $x=1$: $U(1)=(1-3)^2+1=4+1=5.$ Squares in $B$ not exceeding $5$ are $1,4$. So 2 choices.

• $x=2$: $U(2)=(2-3)^2+1=1+1=2.$ Squares $\le 2$: only $1$. So 1 choice.

• $x=3$: $U(3)=0+1=1.$ Squares $\le 1$: only $1$. So 1 choice.

• $x=4$: $U(4)=1+1=2.$ Squares $\le 2$: only $1$. So 1 choice.

• $x=5$: $U(5)=4+1=5.$ Squares $\le 5$: $1,4$. So 2 choices.

• $x=6$: $U(6)=9+1=10.$ Squares $\le 10$: $1,4,9$. So 3 choices.

• $x=7$: $U(7)=16+1=17.$ Squares $\le 17$: $1,4,9,16$. So 4 choices.

• $x=8$: $U(8)=25+1=26.$ Squares $\le 26$: $1,4,9,16,25$. So 5 choices.

• $x=9$: $U(9)=36+1=37.$ Squares $\le 37$: $1,4,9,16,25,36$. So 6 choices.

4. Multiply independent choices

Since choices of $f(x)$ for different $x$ are independent, total number of functions is $2\cdot 1\cdot 1\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6.$ Now, $2\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6=4\cdot 3\cdot 4\cdot 5\cdot 6.$ Compute stepwise: $2\cdot 1\cdot 1\cdot 1\cdot 2=4,$ $4\cdot 3=12,$ $12\cdot 4=48,$ $48\cdot 5=240,$ $240\cdot 6=1440.$ So the number of such functions is $\boxed{1440}.$

5. Compare with stored correct answer

Stored correct answer = $1440$.

Our derived answer also is $1440$, so they agree.