Given $f_n : [0, \infty) \mapsto \mathbb{R}$ a sequence of functions, defined by $$f_{n}(x) = \frac{x}{n^2}\ e^{\frac{-x}{n}} $$
for $n \geq 1$
I am trying to prove that $f_n$ converges uniformly to the zero function by showing that $$\delta_{n} := \text{sup}\{|f_n(x)| : x \in [0, \infty) \} \longrightarrow 0$$ but I don't understand $\delta_{n}$ very well. Does it depend on $n$? How do I actually find it?
You need to show that the supremum tends to $0$, not to necessarily find the supremum exactly. Namely, it is sufficient to find $(a_n)_n$ such that $$ \forall n\geq 1,\qquad \sup_{x\in[0,\infty)} \lvert f_n(x)\rvert \leq a_n $$ and $\lim_{n\to\infty} a_n =0$.
Therefore, let's try and do that here: for all $x\geq 0$ and $n\geq 1$, $$ 0 \leq \frac{x}{n^2} e^{-\frac{x}{n}} = \frac{1}{n}\cdot \frac{x}{n} e^{-\frac{x}{n}} \leq \frac{1}{n}\cdot \sup_{t\in[0,\infty) } t e^{-t} \stackrel{\rm def}{=} a_n $$ (can you see why?) But it is easy to study the function $h(t) = t e^{-t}$: it is differentiable, and $h'(t) = e^{-t}(1-t)$, and it is near-immediate to conclude it achieves its maximum at $t=1$, where $h(1) = 1/e$. Therefore, in light of the above you have $$ \forall n\geq 1,\qquad \sup_{x\in[0,\infty)} \lvert f_n(x)\rvert \leq \frac{1}{en} $$ and you can conclude.
Note: in this case, it so happens that the $a_n$ we found is actually equal to $\sup_{x\in[0,\infty)} \lvert f_n(x)\rvert$.