Denote the transpose, complex conjugate, and hermitian conjugates of $X$ as $(\,X^T,\,X^C,\,X^H),\,$ respectively.
Further, a colon will denote the trace/Frobenius product, i.e. $\,\,A\!:\!B={\rm tr}(A^TB)$
Now consider a single scalar function and its differential
$$\eqalign{
\phi &= z^THw = H:zw^T \cr
d\phi &= H:(dz\,w^T + z\,dw^T) \cr
&= Hw:dz + z^TH:dw^T \cr
&= Hw:dz + H^Tz:dw \cr
}$$
But since
$$ d\phi
= \bigg(\frac{\partial\phi}{\partial z}\bigg)\,dz
+ \bigg(\frac{\partial\phi}{\partial w}\bigg)\,dw
$$
we can identify
$$\eqalign{
&\frac{\partial\phi}{\partial z} = Hw,\,\,\,\,\,&\frac{\partial\phi}{\partial w} = H^Tz \cr
&\frac{\partial\phi}{\partial z^C} = 0,\,\,\,\,\,&\frac{\partial\phi}{\partial w^C} = 0 \cr
}$$
Now substitute $\,z = \{x,x^C\}\,$ and $\,w = \{y,y^C,x,x^C\}\,$ to answer each of your questions.
The gradient looks correct, but the hessian doesn't. Here's how I did the calculations.
In order to write the function in purely matrix form, first note that the $\{x_i\}$ vectors are columns of a single matrix $X$. Next use $(\circ)$ to denote the elementwise/Hadamard product and (:) to denote the trace/Frobenius product, i.e.
$$A:B = {\rm Tr}(A^TB)$$
Define the following variables.
$$\eqalign{
a &= t\circ X^Tw &\implies da = t\circ X^Tdw \cr
b &= \exp(-a) &\implies db = -b\circ da \cr
p &= \exp(a) &\implies dp = p\circ da \implies 1=b\circ p \cr
c &= \log(1+b) &\implies dc = \frac{db}{1+b} \cr
}$$
Write the function in terms of these variables. Then calculate its differential and and back-substitute variables until we arrive at the gradient with respect to $w$.
$$\eqalign{
f &= \mu\,w:w + 1:c \cr
df &= 2\mu\,w:dw + 1:dc \cr
&= 2\mu\,w:dw + \frac{1}{1+b}:db \cr
&= 2\mu\,w:dw - \frac{1}{1+b}:b\circ da \cr
&= 2\mu\,w:dw - \frac{b}{1+b}:t\circ X^Tdw \cr
&= 2\mu\,w:dw - X\Big(\frac{t\circ b}{1+b}\Big):dw \cr
&= \bigg(2\mu\,w - X\Big(\frac{t}{p+1}\Big)\bigg):dw \cr
g = \frac{\partial f}{\partial w} &= 2\mu\,w - X\Big(\frac{t}{1+p}\Big) \cr
}$$
Now find the differential and gradient of $g$.
$$\eqalign{
dg
&= 2\mu\,dw + X\Big(\frac{t\circ dp}{(1+p)\circ(1+p)}\Big) \cr
&= 2\mu\,dw + X\Big(\frac{t\circ p\circ da}{1+2p+p\circ p}\Big) \cr
&= 2\mu\,dw + X\Big(\frac{t\circ p\circ t\circ X^Tdw}{1+2p+p\circ p}\Big) \cr
}$$
Replace the Hadamard products with diagonal matrices, e.g.
$$\eqalign{
P &= {\rm Diag}(p),\,\,
T &= {\rm Diag}(t),\,\,
I &= {\rm Diag}(1) \cr
Px &= p\circ x \cr
}$$
Therefore
$$\eqalign{
dg &= \Big(2\mu I + X(I+2P+P^2)^{-1}T^2PX^T\Big)\,dw \cr
H = \frac{\partial g}{\partial w} &= 2\mu I + X(I+2P+P^2)^{-1}T^2PX^T \cr
}$$
Best Answer
Work on it one piece at a time.
The first piece $$\eqalign{K &= \|X-Y\|_F^2 = (X-Y)^*:(X-Y) \cr dK &= (X-Y)^*:dX}$$ The second piece. $$\eqalign{M &=\|AXe\|_F^2 =(AXe)^*:(AXe)\cr dM &=(AXe)^*:A\,dX\,e =A^TA^*X^*ee^T:dX}$$ And the third. $$\eqalign{ N &={\rm Tr}(X^*BX^T) =X^*B:X\cr dN &=X^*B:dX}$$ Now put it all together, with various summation coefficients (omit the constant terms). $$\eqalign{ L &= K + \sum_iu_iM_i + \sum_kv_kN_k \cr dL &= \Big((X-Y)^* + \sum_iu_iA_i^TA_i^*X^*e_ie_i^T + \sum_kv_kX^*B_k\Big):dX \cr \frac{\partial L}{\partial X} &= X^*-Y^* + \sum_iu_iA_i^TA_i^*X^*e_ie_i^T + \sum_kv_kX^*B_k \cr\cr }$$ In the above derivation, a colon denotes the double-dot product $$A:B = {\rm Tr}(A^TB)$$ Also $X^*$ is treated as being independent of $X$ under differentiation, also known as Wirtinger derivatives or the ${\mathbb {CR}-}$calculus.
And $e_i$ denotes the $i^{th}$ standard basis vector for ${\mathbb R}^{n}$