f∈O(g) means there's N and c>0 such that n>N implies f(n) <= cg(n).

Note that for n>2, n^2 >= 3n.

Then, for n>2, 2(n^2-n) = n^2 - 2n + n^2 >= n^2 - 2n + 3n = n^2 + n.

Summarizing: for n>2, n^2+n <= 2(n^2-n). Thus n^2+n ∈ O(n^2-n)

$f \in O(g)$ means there's $n_0$ and $c>0$ such that $n>n_0$ implies $f(n) \leq cg(n)$.

Note that for $n>2$, we have $n^2 \geq 3n$.

Then, for $n>2$, we have: $$ \begin{align} 2(n^2 - n) & = n^2 - 2n + n^2 \\ & \geq n^2 - 2n + 3n\\ & = n^2 + n\\ \end{align} $$

Summarizing: for $n>2$, we have $n^2+n \leq 2(n^2-n)$. Thus $n^2+n \in O(n^2-n)$.