Find if the polynomials are separable:
a) $\mathbb{Z}_3[x], f(x) = x^6+x^5+x^4+2x^3+2x^2+x+2$
b) $\mathbb{Q}[x], f(x) = x^4-3x^2$
c) $\mathbb{Z}_2[x], f(x) = x^3-1$
d) $\mathbb{Z}_2[x], f(x) = x^6-1$
Wikipedia says that for a polynomial to be separable over $K$, its number of distinct roots is equal to its degree in an algebraic closure of $K$. So,
in a), there are only $3$ possible roots and degree $6$, so this is not separable.
In b), $x^4-3x^2=0 \implies a^2-3a=0 \mbox{ for $a=x^2$ } \implies a=3, a=0 \implies x^2=3, x^2=0 \implies x = \pm \sqrt{3}, x = 0$. There are $3$ possible roots in an algebraic closure of $\mathbb{Q}$, but the polynomial is of 4th degree, so its inseparable
In c), $f(0) = -1 = 2$ so $0$ is not a root, and we should have $3$ roots for it to be inseparable, so I'd say no.
In d), we have less possible roots than degree, so inseparable
Did I do something wrong?
As I stated in the comments, the quick way to test if a polynomial $f \in K[x]$ (where $K$ is a field) is separable is to compute $\gcd(f, f')$. One can see that this is equivalent to your definition of separable as follows.
Fix an algebraic closure $\overline{K}$ and let $d$ be the degree of $f$. To say that $f$ has $d$ distinct roots over $\overline{K}$ means that $f$ has no multiple roots, i.e., $f$ is squarefree. One can show that a root $\alpha$ of $f$ is a multiple root iff $\alpha$ is also a root of $f'$ (the formal derivative of $f$). This is equivalent to saying that $x-\alpha$ is a factor of both $f$ and $f'$, hence also of $\gcd(f,f')$. Thus we see that $f$ is separable iff $\gcd(f,f') = 1$.
Let's see how this applies to your first example. Since $3 = 0$ in $\mathbb{F}_3$ ($=\mathbb{Z}/3\mathbb{Z}$), then $f' = 2 x^4 + x^3 + x + 1$. We compute $\gcd(f,f')$ using the Euclidean algorithm: \begin{align*} f &= (2 x^{2} + x) f' + 2 x^{2} + 2\\ f' &= (x^{2} + 2 x + 2)(2x^2 + 2) + 0 \, . \end{align*} Thus we see that $\gcd(f,f') = 2 x^{2} + 2$, or $x^2 + 1$ if you prefer a monic $\gcd$. This shows that $f$ is not separable: the roots of $x^2+1$ are multiple roots of $f$.
One can see this explicitly by factoring $f$. Using Sage, I found that $f$ splits completely over $\mathbb{F}_3(i) \cong \mathbb{F}_9$, factoring as $$ f = (x + i - 1) (x - i - 1) (x + i)^{2} (x - i)^{2} $$ where $\pm i$ are the roots of $x^2 + 1$.
Sometimes you can determine if $f$ is squarefree by inspection. In (c), $f = x^3 - 1 = (x-1)(x^2 + x + 1)$. Thus $f$ is separable iff $x^2 + x + 1$ is separable. You can show this is the case by again using the Euclidean algorithm to compute $\gcd(f,f')$. Alternatively, $x^2 + x + 1$ is irreducible over $\mathbb{F}_2$ (degree $2$ and no roots) and since $\mathbb{F}_p$ is a perfect field for every prime $p$, then irreducible polynomials over $\mathbb{F}_p$ are always separable.
(d) is easier to see. By the freshman's dream, $$ f = x^6 - 1 = (x^3)^2 - 1 = (x^3 - 1)^2 $$ so $f$ is not squarefree, hence not separable.