Second order Taylor expansion of Frobenius norm

I have the following function

$||{\bf A} - {\bf BC}||^2_F$,

where ${\bf A} \in \mathbb{C}^{m \times n}$, ${\bf B} \in \mathbb{C}^{m \times k}$, and ${\bf C} \in \mathbb{C}^{k \times n}$, which is a convex smoothf unction w.r.t ${\bf B}$. I am looking for second order Taylor expansion of the function w.r.t ${\bf B}$

Basically for the vectors ${\bf x}, {\bf y}$, I can write the second order Taylor expansion in the following form:

$f({\bf y}) \leq f({\bf x}) + \nabla f({\bf x})^H ({\bf y} - {\bf x}) + ({\bf y}-{\bf x})^H \nabla^2 f({\bf x}) ({\bf y}-{\bf x}) \leq f({\bf x}) + \nabla f({\bf x})^H ({\bf y} - {\bf x}) + \frac{L}{2}||{\bf y}-{\bf x}||_2^2$, where $\{{\bf x}, {\bf y}\} \in \mathbb{C}^{n \times 1}$ and L is Lipchitz constant.

Can someone kindly suggest me how to proceed for Taylor expansion of $||{\bf A} - {\bf BC}||^2_F$ ?

If your matrices are complex, then $B^*$ has no derivative wrt $B$; it's the same for your function $f$ (cf. runway44's comment). But it works if your matrices are real.

Lett $f:B\in M_{m,k}(\mathbb{R})\mapsto tr((A-BC)^T(A-BC))$ (the Frobenius norm).

Then $Df_B:H\in M_{m,k}\mapsto -2tr((A-BC)^THC)$ and

$D^2f_B:(H,K)\in (M_{m,k})^2\mapsto 2tr(C^TK^THC).$

We see that $f$ is convex because $D^2f_B(H,H)=2tr(C^TH^THC)\geq 0$.

Since the second derivative does not depend on $B$, the second order Taylor's formula is an equality

$(*)$ $f(B+H)=f(B)-2tr((A-BC)^THC)+tr(C^TH^THC)$.

Of course, $f(B+H)\geq f(B)-2tr((A-BC)^THC)$.

EDIT. Answer to the OP. In the complex case, the equality $(*)$ becomes

$f(B+H)=f(B)-2Re(tr((A-BC)^*HC))+tr(C^*H^*HC)$ (it's a formal equality and not a Taylor's formula).

We deduce that $f(B+H)\geq f(B)-2Re(tr((A-BC)^*HC))$.