P(x) is an odd polynomial. When (x−3) is factored out, the remainder is 6.

Since you asked for a bit of guidance...

In general, if we divide a degree $n$ polynomial $p(x)$ by a degree $m<n$ polynomial $d(x)$, we will end up with a degree $n-m$ polynomial $q(x)$ and a degree $m-1$ (at most) remainder $r(x)$. (This follows from the division algorithm - it might help to perform a few long or synthetic divisions on some polynomials just to play around with this idea.)

We can now express $p$ in terms of $d$, $q$ and $r$ as:

$$\overbrace{p(x)}^{\text{degree }n}=\underbrace{q(x)}_{\text{degree } n-m}\overbrace{d(x)}^{\text{degree }m}+\underbrace{r(x)}_{\text{degree }m-1\text{ (at most)}}$$

In the problem that you present, they give us that when $p(x)$ is divided by $d(x)=x-3$, the remainder is $6$. By the remainder theorem, we can conclude $p(3)=6$.

Since we are told that $p$ is an odd function, $p(-3)=-p(3)=-6$.

We are asked for the remainder when $p(x)$ is divided by $x^2-9=(x+3)(x-3)$. Notice that this is a degree $2$ polynomial, so the remainder will be a degree $1$ (linear) polynomial. In other words, $r(x)$ will have the form $ax+b$.

Expressing $p(x)$ in terms of this new information, then:

$$p(x)=q(x)(x+3)(x-3)+(ax+b)$$

where $q$ is some polynomial.

From this, we can see that $p(3)=3a+b$, which we know is $6$, and $p(-3)=-3a+b$, which we know is $-6$. Solving these simultaneously gives us $a=2$ and $b=0$. Thus the remainder is the degree $1$ polynomial $2x$.

Hint: by euclidian division $p(x) = (x^2-9) q(x) + a x + b$.

Then $p(3)-p(-3)= 6a = 6 - (-6) = 12$ so $a=2$.

Since $p(x)$ is odd, all coefficients of even powers of $x$ are $0$, including the free term, so $b=0$.

So the remainder is $\cdots$