Let there are complex matrices given by $A \in C^{m×n}$, $B \in C^{n \times s}$ and $C \in \mathbb{C}^{m \times s}$ and I have $f(A)= \|AB - C\|_F^2$ which has a gradient given by $g(A) = (AB - C)B^H$ w.r.t $A$, then as the $f$ has a Lipschitz continous gradient, I have obatined the Lipschitz constant as
$L = \|BB^H\|_F$.
Can someone suggest if it is the correct answer or I am making a mistake?
$\newcommand{\R}{\mathbb{R}}$We say that a function $f:D\to\R^m$, where $D\subseteq \R^n$, is Lipschitz continuous on $D$ with Lipschitz constant $L\geq 0$ if $$ \|f(x) - f(y)\| \leq L \|x-y\|, $$ for all $x, y \in D$.
What is important to note here is that $L$ does not depend on $x$ or $y$.
If $f$ is a $C^1$ function (your function is, indeed, $C^1$), then the best Lipschitz constant on a set $X\subseteq D$ is given by
$$ L = \sup_{x\in X}\|\nabla f(x)\|, $$
but in your case $\nabla f(x)$ is not bounded on $\R^n$, therefore, $f$ is not Lipschitz on $\R^n$, but it is Lipschitz on any compact set $K\subset\subset\R^n$. This property is known as local Lipschitz continuity.
In order to understand this intuitively, think of $f(x) = x^2$ with $x\in\R$. This is locally Lipschitz, but bot globally (see this thread for reference).