I have read a few answers to similar questions on here but I can't get my head around this. This is my attempt to far:
$f(t)$ is monic and irreducible over $\mathbb{Q}$ since it is $2$-Eisenstein. Since $\mathbb{Q}$ has characteristic $0$, $f(t)$ has $3$ different roots, so $\varphi:\text{Gal}(f)\to S_{3}$ is injective. So there is a tower of extensions $\mathbb{Q}(f)\supset\mathbb{Q}(\alpha)\supset\mathbb{Q}$ where $\alpha$ is a root of $f$.
Now, $[\mathbb{Q}(f):\mathbb{Q}]$ is divisible by $[\mathbb{Q}(\alpha):\mathbb{Q}]=\deg(f)=3$, so $|\text{Gal}(f)|$ is divisible by $3$ and so $\text{Gal}(f)$ contains a $3$-cycle.
Is what I have done so far correct? How do I proceed? My gut is telling me the $\text{Gal}(f)\cong S_{3}$ but I don't know how to show it.
All of what you did is solid. To finish note that, in general, the Galois group of an irreducible cubic is $A_3$ iff the discriminant is a square (because $\Bbb Q(\sqrt{\Delta_f})$ is a subfield of your splitting field). In your case $\Delta_f=148$ is not a square, so you get that the full Galois group is $S_3$.