What is the easiest way to prove that a given function is one-to-one? Depends on the function!
For some functions, the definition is simplest: show that if you have $a$ and $b$ in the domain such that $f(a)=f(b)$, then you can conclude that $a=b$.
For some functions, the contrapositive of the definition is simplest: show that if $a$ and $b$ are in the domain and $a\neq b$, then $f(a)\neq f(b)$.
For yet other functions, it is easier to prove they are one-to-one by showing that they have a left inverse: that is, if $f\colon X\to Y$, then coming up with a function $g\colon Y\to X$ such that $g\circ f = \mathrm{id}_X$, the identity of $X$; this immediately implies that $f$ is one-to-one.
When the sets have other structure, there may be other tools available (showing that a linear transformation is one-to-one can be done by showing that if $T(\mathbf{x})=\mathbf{0}$, then $\mathbf{x}=\mathbf{0}$, for example).
Which one is easiest? Depends on the function.
What's the easiest way to show that a function is not one-to-one? The absolutely, definitely best way is to exhibit two specific elements in the domain, $a$ and $b$, showing that $a\neq b$, and then showing that $f(a)\neq f(b)$. In fact, I would say that 99% of the "proofs that a function is not one-to-one" should at the very list end with such an explicit example.
How do you find such examples? Again, it depends on the function. Sometimes you try to see if the function is one-to-one by considering what you can conclude if $f(a)=f(b)$; this will lead you to a relationship between $a$ and $b$, and often you can see that the relation can be satisfied with $a\neq b$. This will lead you to an example. For instance, to show that $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^2$ is not one-to-one, you could go through the following thought process:
"I wonder if it is one to one... let's see... if $f(a)=f(b)$, then this means that $a^2=b^2$; taking square roots, that says that $|a|=|b|$... ah! but that can happen even if $a\neq b$! For instance, if $a=-b$... Okay, so here's a proof that $f$ is not one-to-one: take $a=1$, $b=-1$. Then $a\neq b$, and $f(a)=f(1) = 1$, $f(b)=f(-1) = 1$. So $f(a)=f(b)$, and this shows that $f$ is not one-to-one."
Which one is easiest? Depends on the function.