Why there is no $y_i$ term in $\frac{d^{2} J(\boldsymbol{\alpha})}{d \alpha_{i}^{2}}$? 
$$\frac{d^{2} J(\boldsymbol{\alpha})}{d \alpha_{i}^{2}}=\lambda^{-1} \mathbf{x}_{i}^{\mathrm{T}} \mathbf{x}_{i}+\frac{1}{\alpha_{i}\left(1-\alpha_{i}\right)}$$

I cannot understand why there is no $y_i$ term in $\frac{d^{2} J(\boldsymbol{\alpha})}{d \alpha_{i}^{2}}.$

$$\mathbf{w}(\boldsymbol{\alpha})=\lambda^{-1} \sum_{i} \alpha_{i} y_{i} \mathbf{x}_{i}$$


$$J(\alpha)=\frac{1}{2 \lambda} \sum_{i j} \alpha_{i} \alpha_{j} y_{i} y_{j} \mathbf{x}_{j}^{\mathrm{T}} \mathbf{x}_{i}-\sum_{i} H\left(\alpha_{i}\right)$$


\begin{aligned}
\frac{d J(\boldsymbol{\alpha})}{d \alpha_{i}} &=\lambda^{-1} y_{i} \sum_{j} \alpha_{j} y_{j} \mathbf{x}_{j}^{\mathrm{T}} \mathbf{x}_{i}+\log \frac{\alpha_{i}}{1-\alpha_{i}} \\
&=y_{i} \mathbf{w}(\boldsymbol{\alpha})^{\mathrm{T}} \mathbf{x}_{i}+\log \frac{\alpha_{i}}{1-\alpha_{i}}
\end{aligned}

 A: $
\def\J{{\cal J}}
\def\a{\alpha}\def\b{\beta}\def\e{\varepsilon}\def\l{\lambda}
\def\o{{\tt1}}\def\p{\partial}
\def\L{\left}\def\R{\right}
\def\LR#1{\L(#1\R)}
\def\BR#1{\Big(#1\Big)}
\def\diag#1{\operatorname{diag}\LR{#1}}
\def\Diag#1{\operatorname{Diag}\LR{#1}}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\hess#1#2#3{\frac{\p^2 #1}{\p #2\,\p #3}}
\def\c#1{\color{red}{#1}}
$The Frobenius product is a concise notation for the trace
$$\eqalign{
A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A^TB} \\
A:A &= \big\|A\big\|^2_F \\
}$$
This is also called the double-dot or double contraction product.
When applied to vectors $(n=\o)$ it reduces to the standard dot product.
We are given vectors $(a,y)$ with components $(\a_i,y_i)$ and a matrix $(X)$ with columns $(x_i)$.
In addition, it will prove convenient to define two diagonal matrices and a vector
$$\eqalign{
A = \Diag{a}, \quad Y = \Diag{y},\quad b=\LR{y\odot a}=Ya \\
}$$
where $(y\odot a)$ denotes elementwise multiplication.
Similarly, $\LR{\frac ba}$ will be used to denote elementwise division.
The entropy is defined as
$$\eqalign{
-H(a) = a:\log(a) + (\o-a):\log(\o-a)  \\
}$$
where the log() function is applied elementwise.
Now the objective function can be written in matrix notation without explicit summations
$$\eqalign{
w &= \l^{-1}Xb \;=\; \l^{-1}XYa \\
\J &= \frac{\l}{2}\LR{w^Tw} - H(a) 
   \;\;\doteq\;\; \frac{\l}{2}\LR{w:w}
       + a:\log(a) + (\o-a):\log(\o-a) \\
}$$
Calculate the differential and the gradient of the function.
$$\eqalign{
d\J &= \l w:dw + \log\LR{\frac{a}{\o-a}}:da \\
  &= \LR{XYa}:\LR{\l^{-1}XY\,da} + \log\LR{\frac{a}{\o-a}}:da \\
  &= \l^{-1}{YX^TXYa}:da + \log(a):da - \log(\o-a):da \\
\grad{\J}{a}
  &= \l^{-1}{YX^TXYa} + \log(a) - \log(\o-a)
 \;\doteq\; g \qquad\big({\rm the\;gradient}\big) \\
}$$
Now calculate the differential and the gradient of the gradient.
$$\eqalign{
dg &= \l^{-1}{YX^TXY\,da} + \frac{da}{a} - \frac{d(\o-a)}{\o-a} \\
  &= \l^{-1}\LR{YX^TXY}\,da + \LR{A-A^2}^{-1}da \\
\grad{g}{a} 
  &= \l^{-1}\BR{\Diag{y}\;X^TX\;\Diag{y}} + \LR{A-A^2}^{-1}
 \;\doteq\; \hess{\J}{a}{a^T} \\
}$$
Therefore, $y$-terms are present in the Hessian.
However, if you take one more derivative (i.e the gradient of the Hessian) then the $y$-terms will be annihilated.
