I know I've asked for the umpteenth time on taking partial derivatives to solutions of elliptic PDE. But I have yet one more.
Page 351 of PDE Evans:
Proof. 1. We may assume $u > 0$ in $U$, for otherwise we could apply the result to $u + \epsilon$ and then let $\epsilon \rightarrow 0^+$.
$\quad$Set $$v := \log u \tag{19}$$ Since $Lu = 0$, we compute $$\sum_{i,j=1}^n a^{ij} v_{x_i x_j} + a^{ij} v_{x_i}v_{x_j}=0 \quad \text{in }U. \tag{20}$$ Define $$w := \sum_{i,j=1}^n a^{ij} v_{x_i} v_{x_j},$$ so that $\text{(20)}$ says $$-\sum_{i,j=1}^n a^{ij} v_{x_i x_j} = w. \tag{22}$$
$\quad$2. We calculate for $k,l = 1,\ldots,n$ that $$w_{x_k x_l} = \sum_{i,j=1}^n 2a^{ij} v_{x_i x_k x_l} v_{x_j} + 2a^{ij} v_{x_i x_k} v_{x_j x_l}+R,$$ where the remainder term $R$, resulting from derivatives falling upon the coefficients, satisfies an estimate of the form $$|R| \le \epsilon |D^2 v|^2 + C(\epsilon) |Dv|^2 \tag{23}$$ for each $\epsilon > 0$.
Given $w:=\sum_{i,j=1}^n a^{ij} v_{x_i} v_{x_j}$, I am trying to derive $$w_{x_k x_l} = \sum_{i,j=1}^n 2a^{ij} v_{x_i x_k x_l} v_{x_j} + 2a^{ij} v_{x_i x_k} v_{x_j x_l}+R.$$
I tried differentiating $w$ once first, with respect to $x_k$. Then I differentiated $w$ with respect to $x_l$. However, I did not arrive at the same result for $w_{x_k x_l}$. In fact, I don't know how they get the remainder term of $R$ to begin with.
As you guys may have noticed, there are also two expressions of $w$: one defined by $\text{(21)}$ and one by $\text{(22)}$ (the latter of which is aided by $\text{(20)}$. Somehow both of these were used to calculate the expression for $w_{x_k x_l}$.
Use the product rule term-by-term to find $$w_{x_k} = \sum_{i,j} a^{ij}v_{x_i x_k}v_{x_j} + a^{ij}v_{x_i}v_{x_jx_k} + a^{ij}_{x_k}v_{x_i}v_{x_j} $$ Then do the same type of thing again. All of the terms where a derivative fall on $a^{ij}$ go into $R$. We get $$w_{x_kx_l} = \sum_{i,j} a^{ij} (v_{x_i x_k x_l}v_{x_j} +v_{x_i x_k}v_{x_jx_l} +v_{x_i x_l}v_{x_jx_k} +v_{x_i}v_{x_jx_kx_l}) +R$$
Now you need to play with the sum indices: on the first two terms in the sum, flip the $i$ and $j$. Then you can combine the terms to get the $2$'s in the expression. Added in edit: this last step assumes $a^{ij} = a^{ji}$, as pointed out by Semiclassical. This hypothesis appears in the middle of page 344.