I have the following Lyapunov function:
$$V(x,y)=\frac{1}{2}y^2+\int_0^xh(s)\text{ d}s$$
I am given that $x\cdot h(x) >0$ if $x\neq 0$ and that $H(x)=\int_0^xh(s)\text{ d}s\rightarrow\infty\text{ as }|x|\rightarrow\infty$. h is $C^0$.
How does this guarantee that V(x,y) is positive definite everywhere except (0,0) (the equilibrium point in question)?
As you said, for nonzero $x$ we have $x\,h(x)>0$. Therefore, if $x>0$ then $h(x)>0$ and you can see that $$\int_0^x h(s)ds>0\tag{1}$$ Because the sum of positive values is positive and sum of negative values is negative. So if $x<0$ then $h(x)<0$ and $\int_x^0 h(s)ds<0$. This implies $(1)$ for all nonzero values of $x$.
On the other hand, $$V(0,0)=0+\int_0^0h=0$$ and $$V(\infty)=\lim_{x\text{ or }y\rightarrow\infty} \frac 12y^2+\int_0^x h(s)ds=\infty$$ So on everywhere except the origin, $V$ is positive. It is also radially unbounded and vanishes at the origin, which implies $V$ is a Lyapunov function.