I don't think your "result so far" is entirely accurate.
Defining $\mathbb I=$ (matrix of all ones), $\,E=({\mathbb I}-I)$,
and using the Hadamard ($\circ$) product, we can use $(-E\circ X)$ to replace those bulky $({\rm diag}(X)-X)$ terms.
Ignoring the L1-term, and using the Frobenius (:) product, the function to be minimized is
$$
f = \frac{\lambda}{2}(X):(X)-(\Delta):(E\circ X) + \frac{\rho}{2}(K-E\circ X):(K-E\circ X)
$$
The differential of which is
$$\eqalign{
df &= (\lambda X):(dX)-(\Delta):(E\circ dX)-\rho(K-E\circ X):(E\circ dX) \cr
&= (\lambda X):(dX)-(\Delta\circ E):(dX)-\rho(E\circ K-E\circ E\circ X):(dX) \cr
&= [\lambda X-\Delta\circ E-\rho E\circ(K-X)]:(dX) \cr
}$$
Since $df=\frac{\partial f}{\partial X}:dX,\,$ the derivative must be
$$\eqalign{
\frac{\partial f}{\partial X} &= \lambda X-E\circ\Delta-\rho E\circ(K-X) \cr
&= \lambda X+({\rm diag}(\Delta)-\Delta)+\rho({\rm diag}(K)-K+X-{\rm diag}(X))\cr
}$$
Which is different from your result -- unless you're utilizing special properties of {$X,\Delta$} which you forgot to tell us about, in order to simplify the result.
Update #1
Since you're using $\|X\|_2$ to denote the Frobenius norm, I assume you are using Schatten norms rather than induced norms.
In that case, $\|X\|_1$ denotes the Nuclear norm, for which the derivative is known.
$$ \eqalign{
\frac{\partial\,\|X\|_{*}}{\partial X} &= X(X^TX)^{-1/2} \cr
}$$
Update #2
Upon further reflection, I suppose you could be using entrywise norms, in which case $\|X\|_2$ still denotes the Frobenius norm, but $\|X\|_1$ denotes the Manhattan norm.
$$ \eqalign{
\|X\|_1 &= {\mathbb I}:{\rm abs}(X) \cr
d\,\|X\|_1 &= {\mathbb I}:d({\rm abs}(X)) \cr
&= {\mathbb I}:({\rm sign}(X)\circ dX) \cr
&= ({\mathbb I}\circ{\rm sign}(X)):dX \cr
&= {\rm sign}(X):dX \cr
\frac{\partial\,\|X\|_1}{\partial X} &= {\rm sign}(X) \cr
}$$
where the {${\rm abs},{\rm sign}$} functions are applied entrywise as well.
