On Wikipedia the de Rham cohomology groups are defined to be the cohomology groups of the de Rham cochain complex (equivalence classes of differential $k$-forms).
By this definition the zeroth de Rham cohomology group is the set of all closed differential zero forms modulo all exact $0$-forms (i.e. modulo the image of the exterior derivative). In formula,
$$ H^0_{dR} = {\ker d^{1}\over \mathrm{im } d^0 } = \ker d^{1}$$
Since $d^0: 0 \to \Omega^0$ is the trivial map.
Question1: Am I correct so far?
Using the notation and terminology on Wikipedia $H_0$ is therefore the set of all closed $0$-forms. Since $0$-forms are smooth functions the question arises what it means for a smooth function $f$ to be closed, that is, which $f$ have vanishing exterior derivative $df=0$.
Question2: How to determine whether a smooth function is closed?