= Solution
An <elastic deformation> is reversible: after removal of the deforming loads the material returns to its reference state, and its instantaneous <stress> is determined by its current deformation rather than a plastic or viscous history. In the infinitesimal approximation the <displacement field> $u_i$ gives the <infinitesimal strain tensor> $e_{ij}=(u_{i,j}+u_{j,i})/2$. The <Cauchy stress tensor> is defined by <traction> $t_i=\sigma_{ij}n_j$ on a plane of unit normal $\mathbf n$; absence of body couples makes $\sigma_{ij}=\sigma_{ji}$.
Let $\mathcal E(e)$ be the recoverable <elastic energy> per unit mass. The local mechanical power per unit volume is
$$
\sigma_{ij}\partial_j\dot u_i=\sigma_{ij}\dot e_{ij},
$$
because the symmetric <Cauchy stress tensor> annihilates the antisymmetric part of the velocity gradient. Conservation of <energy> during the deformation gives $\rho\dot{\mathcal E}=\sigma_{ij}\dot e_{ij}$. Since $\dot{\mathcal E}=(\partial\mathcal E/\partial e_{ij})\dot e_{ij}$ for arbitrary symmetric <strain> rates,
$$
\boxed{\sigma_{ij}=\rho\frac{\partial\mathcal E}{\partial e_{ij}}.}
$$
Here the derivative is the gradient on the space of symmetric tensors, with the double-contraction inner product. This convention accounts for the two occurrences of each off-diagonal entry; treating six independent coordinates without their corresponding factors would give an erroneous factor of two in shear components.
Take a stress-free reference state of <mass density> $\rho_0$. The <Taylor series> of the <elastic energy> starts quadratically:
$$
\mathcal E(e)=\mathcal E(0)+\frac1{2\rho_0}c_{ijkl}e_{ij}e_{kl}+O(|e|^3).
$$
Using $\rho=\rho_0+O(|e|)$ gives $\sigma_{ij}=c_{ijkl}e_{kl}+O(|e|^2)$, the relation of <linear elasticity>. If the reference state has initial <stress>, the first-order expansion also retains that constant <stress>. Symmetry of <stress> and <strain> gives the minor symmetries $c_{ijkl}=c_{jikl}=c_{ijlk}$, while equality of the mixed second derivatives of <elastic energy> gives the major symmetry $c_{ijkl}=c_{klij}$. Thus the <elastic stiffness tensor> is a symmetric map on a six-dimensional <strain> space: \b[there are $6\cdot7/2=21$ independent coefficients].
For <twofold elastic symmetry>, choose the symmetry axis as $x_3$. Rotation through $\pi$ is $Q=\operatorname{diag}(-1,-1,1)$, and transforms the <infinitesimal strain tensor> into $QeQ^T$. Components $e_{11},e_{22},e_{33},e_{12}$ are even and $e_{23},e_{13}$ are odd. Invariance of <elastic energy> excludes every coupling between these parity classes. In the engineering <strain> ordering $(e_{11},e_{22},e_{33},2e_{23},2e_{13},2e_{12})$, with corresponding <stress> ordering $(\sigma_{11},\sigma_{22},\sigma_{33},\sigma_{23},\sigma_{13},\sigma_{12})$, the symmetric stiffness matrix has the form
$$
C=\begin{pmatrix}
C_{11}&C_{12}&C_{13}&0&0&C_{16}\\
C_{12}&C_{22}&C_{23}&0&0&C_{26}\\
C_{13}&C_{23}&C_{33}&0&0&C_{36}\\
0&0&0&C_{44}&C_{45}&0\\
0&0&0&C_{45}&C_{55}&0\\
C_{16}&C_{26}&C_{36}&0&0&C_{66}
\end{pmatrix}.
$$
Its symmetric even block has ten independent entries and its symmetric odd block has three. Hence \b[a twofold axis leaves $13$ independent coefficients]. No invariance under other rotation angles has been assumed.
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