Derivating matrixs. I don't know how to derive respect to $z$ this 
$$\frac{\beta}{2} \parallel x-Hz\parallel^{2}+\frac{\alpha}{2} z^{t}C^{t}Cz$$
where $z$ and $x$ are n-dimensional vectors and $H$ and $C$ matrix of order $n\times n$.
The result is $(\alpha C^{t}C+\beta H^{t}H)z-\beta H^{t} x$ but I don't know how to get to it.
 A: Using the product notation (:) for the trace, i.e.
$$A:B={\rm tr}(A^TB)$$
allows you to write the objective function in a nicer form.
The differential and gradient then follow easily. 
$$\eqalign{
 \phi &= \frac{\beta}{2}(Hz-x):(Hz-x) + \frac{\alpha}{2}(Cz):(Cz)  \cr
d\phi &= \beta(Hz-x):H\,dz + \alpha(Cz):C\,dz  \cr
  &= \big(\beta H^T(Hz-x) + \alpha C^TCz\big):dz  \cr
\frac{\partial\phi}{\partial z} &= \beta H^T(Hz-x) + \alpha C^TCz \cr\cr
}$$
A: Not a proof but some "intuition":
In one dimension,
$$\frac{d}{dz}\left(\frac{\beta}{2} (x - h z)^2 + \frac{\alpha}{2} c^2 z^2\right) = h \beta (zh-x) + \alpha c^2.$$
This resembles the answer in higher dimensions, and the computation may be similar.

Slightly more rigorous. Here are some facts that you can check directly by computing partial derivatives.

If $f:\mathbb{R}^n \to \mathbb{R}$ and $f(y) = \|y-x\|^2$, then $\nabla f(y) = 2 y$.
If $f:\mathbb{R}^n \to \mathbb{R}^n$ and $f(z) = Hz$ then $J_f(z) = H$.
If $f:\mathbb{R}^n \to \mathbb{R}$ and $f(z) = z^t C^t C z$ then $\nabla f(z) = 2 C^t C$.

It turns out you can stitch these together using a sort of chain rule to arrive at your answer.

Bare hands / direct method. Just take the partial derivative of your expression with respect to each $z_i$ and stack all the partial derivatives into a vector.
A: We have 
\begin{align}
\frac{\beta}{2}  \|x-Hz\|^2+\frac{\alpha}{2} z^{t}C^{t}Cz
=&
\frac{\beta}{2} (x-Hz)^T(x-Hz)+\frac{\alpha}{2}(Cz)^{t}Cz=
\\
=
\frac{\beta}{2} 
\left (
\left\lgroup
\begin{array}{c}
x_1\\
x_2\\
\vdots\\
x_n
\end{array}
\right\rgroup
-
\left\lgroup
\begin{array}{cccc}
H_{11}&H_{12}&\ldots&H_{1n}\\
H_{21}&H_{22}&\ldots&H_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
H_{n1}&H_{n2}&\ldots&H_{nn}\\
\end{array}
\right\rgroup
\cdot 
\left\lgroup
\begin{array}{c}
z_1\\
z_2\\
\vdots\\
z_n
\end{array}
\right\rgroup
\right)^T
&
\left (
\left\lgroup
\begin{array}{c}
x_1\\
x_2\\
\vdots\\
x_n
\end{array}
\right\rgroup
-
\left\lgroup
\begin{array}{cccc}
H_{11}&H_{12}&\ldots&H_{1n}\\
H_{21}&H_{22}&\ldots&H_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
H_{n1}&H_{n2}&\ldots&H_{nn}\\
\end{array}
\right\rgroup
\cdot 
\left\lgroup
\begin{array}{c}
z_1\\
z_2\\
\vdots\\
z_n
\end{array}
\right\rgroup
\right)
\\
+\frac{\alpha}{2}
\left(
\left\lgroup
\begin{array}{cccc}
C_{11}&C_{12}&\ldots&C_{1n}\\
C_{21}&C_{22}&\ldots&C_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
C_{n1}&C_{n2}&\ldots&C_{nn}\\
\end{array}
\right\rgroup
\cdot 
\left\lgroup
\begin{array}{c}
z_1\\
z_2\\
\vdots\\
z_n
\end{array}
\right\rgroup
\right)^T
&
\left(
\left\lgroup
\begin{array}{cccc}
C_{11}&C_{12}&\ldots&C_{1n}\\
C_{21}&C_{22}&\ldots&C_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
C_{n1}&C_{n2}&\ldots&C_{nn}\\
\end{array}
\right\rgroup
\cdot 
\left\lgroup
\begin{array}{c}
z_1\\
z_2\\
\vdots\\
z_n
\end{array}
\right\rgroup
\right)
=
\\
=\frac{\beta}{2}
\left(
x_j-\sum_{k=1}^{n}(H_{kj}z_k
\right)^T
\cdot 
\left(
x_j-\sum_{k=1}^{n}(H_{kj}z_k
\right)
+&
\frac{\alpha}{2}
\left(
\sum_{\ell=1}^{n}C_{i\ell}z_{\ell}
\right)^T
\left(
\sum_{\ell=1}^{n}C_{i\ell}z_{\ell}
\right)
\\
=
\frac{\beta}{2}\sum_{j=1}^{n}x_j^2
-
{\beta}\sum_{j=1}^{n}\sum_{k=1}^{n}x_jH_{jk}z_k
+&
\frac{\beta}{2}\sum_{j=1}^{n}\sum_{k=1}^{n}H_{jk}^2z_k^2
+
\frac{\alpha}{2}\sum_{i=1}^{n}\sum_{\ell=1}^{n}C_{i\ell}z_{\ell}
\end{align}
For $u=1,2,3,\ldots, n$ we have 
$$
\frac{\partial}{\partial z_u}
\left( 
\frac{\beta}{2}  \|x-Hz\|^2+\frac{\alpha}{2} z^{t}C^{t}Cz
\right)
=\\
=
-
{\beta}\sum_{j=1}^{n}x_jH_{ju}z_u
+
\frac{\beta}{2}\sum_{j=1}^{n}H_{ju}^2z_u^2
+
\frac{\alpha}{2}\sum_{i=1}^{n}C_{iu}z_{u}
$$
