Being a lazy sort, I typically use Iverson brackets for the purpose. Recall that $[p]$ is $1$ if condition $p$ is true, and $0$ if condition $p$ is false.
With this, we can write your integral as
$$\iint [0\leq x\leq 1][0\leq y\leq 3x]f(x,y)\,\mathrm dy\mathrm dx=\iint [0\leq x\leq 1]\left[0\leq \frac{y}{3}\leq x\right]f(x,y)\,\mathrm dy\mathrm dx$$
This can be treated as an integral with doubly infinite limits; the Iverson brackets zero things out outside the domain of validity.
Now, Iverson brackets have the property that $[p\text{ and }q]=[p][q]$; we can use this property to give the alternate representation
$$\iint \left[0\leq \frac{y}{3}\leq x\leq 1\right]f(x,y)\,\mathrm dx\mathrm dy$$
where I have already taken the liberty to swap out the differentials.
Now, we can factor the Iverson bracket as
$$\iint \left[0\leq \frac{y}{3}\leq 1\right]\left[\frac{y}{3}\leq x\leq 1\right]f(x,y)\,\mathrm dx\mathrm dy$$
or
$$\iint \left[0\leq y\leq 3\right]\left[\frac{y}{3}\leq x\leq 1\right]f(x,y)\,\mathrm dx\mathrm dy$$
You can then translate this back into the usual notation:
$$\int_0^3\int_{y/3}^1 f(x,y)\,\mathrm dx\mathrm dy$$
Best Answer
In general you cannot switch the order of integration without additional constraints. These are typically given by Fubini's theorem. In particular the example you've given does not converge absolutely so switching the order changes the answer.