A field \(k\) is perfect if and only if either \(\ch k=0\), or \(\ch k=p>0\) and the Frobenius map \(F\colon k\to k\), \(F(a)=a^p\), is surjective, hence an automorphism. In particular, every field of characteristic zero and every finite field is perfect.
An irreducible \(f\in k[x]\) has a repeated root if and only if \(\gcd(f,f')\ne1\), and since \(f\) is irreducible and \(\deg f'<\deg f\), this happens if and only if \(f'=0\).
If \(\ch k=0\), then \(f'\ne0\) for every nonconstant \(f\), so \(k\) is perfect.
Let \(\ch k=p>0\). Then \(f'=0\) if and only if \(f(x)=\sum_i a_ix^{pi}\) for some \(a_i\in k\). If \(F\) is surjective, choose \(b_i\in k\) with \(b_i^p=a_i\); then \(f=\qty{\sum_ib_ix^i}^p\) is reducible, so no irreducible polynomial has \(f'=0\), and \(k\) is perfect. If \(F\) is not surjective, choose \(a\in k\) that is not a \(p\)th power, and let \(\beta\) be a root of \(x^p-a\) in a splitting field, so \(x^p-a=(x-\beta)^p\). A monic factor of \(x^p-a\) in \(k[x]\) of degree \(j\) with \(0<j<p\) is \((x-\beta)^j\), whose coefficient of \(x^{j-1}\) is \(-j\beta\); since \(j\) is a unit in \(k\), this would give \(\beta\in k\) and \(a=\beta^p\). So \(x^p-a\) is irreducible and not separable, and \(k\) is not perfect.
The Frobenius map is an injective ring homomorphism, so on a finite field it is surjective.