Consider the group of positive real numbers $(\mathbb{R}>0)$ acting on the real numbers by multiplication. What is the stabiliser subgroup of $x=0$?
What I've tried so far
So as I know, the stabiliser subgroup of an element $x$ is the set of all elements that leave $x$ invariant. The element $x=0$ can be multiplied by any real number to give $0$ (itself). So would this mean that the stabiliser subgroup for $x=0$ in the Group in my case is the set of all real numbers $\mathbb{R}$?
Your reasoning is definitely correct until the last statement. Stabilizer of $0$ is not $\mathbb{R}$ but it should be $\mathbb{R}^+$. Note that if a group $G$ acts on a set $A$, then stabilizer of $a \in A$ in $G$ is $$G_a = \{g\in G\ |\ g \cdot a = a\}$$ So, the elements that fix $a$ are all in the group $G$. Also note that $G_a \le G, \forall a \in A$.