Question: I am trying to find the dimension of the vector space ($P\_2$ means polynomials of degree at most $2$) 
$$V = \{p(x) \in P\_2 \mid xp'(x) = p(x)\}.$$ 
However, I don't even know how to start...

(Additionally, the question asks to find the basis for $V$, which I don't know how to do, but I think I might if someone tells me how to understand the above first...)

Answer: 
Consider the linear transformation $T\colon P\_2\to P\_2$ given by $T(p) = xp' - p$. You should verify that this is indeed a linear transformation.

For example, if $p(x) = 3x^2 - 2x$, then $p'(x) = 6x-2$, so $xp'(x) = x(6x-2) = 6x^2-2x$. Therefore,
$$T(p) = xp' - p = (6x^2-2x)-(3x^2-2x) = 6x^2-2x-3x^2+2x = 3x^2.$$

The vector space you want is nothing more than the nullspace of this linear transformation. So you can compute the nullspace with any method you may already know (e.g., find a basis for $P\_2$, find the coordinate matrix for $T$ relative to that basis, and then find the nullspace of the matrix), and use it to find its dimension.
Taking an arbitrary element $p(x)=ax^2+bx+c\in V$, we see it has to satisfy the special condition $xp'(x)=p(x)$, so $x(2ax+b)=2ax^2+bx+0=ax^2+bx+c$.

Then immediately $c=0$, and $2a=a$ implies that $a=0$.

So the only things in $V$ are of the form $bx$ for any $b\in F$.

I think you can work out a basis :)