Combinatorics Recurrence relation 
Let $h_n$ be a number sequence where $h_n = 3h_{n-1} - 2h_{n-2}$ with $h_0 = 0$ and $h_1 = 1$. Compute the ordinary generating function of $h_n$ and then using the generating function compute a formula for $h_n$. 

Does this start looks right?
We write the recurrence relation in form: 
$$h_n - 3h_{(n-1)} + 2h_{(n-2)} = 0$$ 
Let $g(x) = h_0 + h_1 x + h_2 x^2+\ldots$ be the generating function for sequence $h_0, h_1,\ldots$.
We have $g(x) = h_0 + h_1 x + \ldots$ as well as $-3x\,g(x) = -3h_0 x -3 h_1 x^2 - \ldots$ and  $2x^2\,g(x) = 2h_0 x^2 + 2h_1 x^3 +\ldots$.
Adding all three would give us :
$$(1 -3x + 2x^2)  g(x) = h_0 + (-3h_0 + h_1) x + (2h_0 -3h_1 + h_2) x^2 + \ldots + (h_n -3h_{n-1} + h_{n-2}) x^n + \ldots$$
Since $h_n - 3h_{n-1} + 2 h_{n-2} = 0$ and since $h_0 = 0, h_1 = 1$.
So we know that $(1 - 3x + 2x^2) g(x)= 1 + x$ and that leads to:
$$ g(x) = \frac{1+x}{(1-2x)(1-x)}.$$ 
Does this look like a correct start?

Once you have:
$$ g(x)=\sum_{n\geq 0}h_n x^n = \frac{\color{red}{x}}{(1-2x)(1-x)} = \frac{1}{1-2x}-\frac{1}{1-x}\tag{1}$$
it is straightfoward to check that $\color{red}{h_n=2^n-1}.$ 
Just check your computations since the value in zero of $\frac{1+x}{(1-2x)(1-x)}$ is one, while you have $h_0=0$.
