As the language
$$ L_1 = \{ a^ib^j \mid 1 \le i, 1\le j, \;ij < 10\} $$
is finite, it is regular. Moreover
$$ L_2 = \{a^ib^j \mid i,j \ge 1 \} $$
is regular (in has $aa^*bb^*$ as regular expression). But so $L = L_2 \setminus L_1 = L_2 \cap \complement L_1$ as the set of regular languages is closed under taking complements and finite intersections.
Edit: Regarding Nishant's questions. We have by the definition of complement and intersection
\begin{align*}
L &= \{a^n b^m \mid nm \ge 10\}\\
&= \{a^n b^m \mid \neg(nm < 10)\}\\
&= \{w \in \Sigma^* \mid \exists n,m\ge 1 : w = a^nb^m \wedge \neg(nm < 10)\}\\
&= \{w \in \Sigma^* \mid \exists n,m \ge 1 : w = a^nb^m\} \cap \{w \in \Sigma^* \mid \neg\exists n,m \ge 1: w=a^nb^m, nm < 10\}\\
&= L_2 \cap \complement L_1
\end{align*}