= Solution
For a fundamental <electric field> in the $xy$ plane, $E_z=0$. The specified <tensor> then gives
$$
P_x=P_y=0,\qquad
P_z(2\omega)=2\epsilon_0d_{36}E_xE_y.
$$
Writing $E_x=E_0\cos\phi$ and $E_y=E_0\sin\phi$ yields
$$
P_z=\epsilon_0d_{36}E_0^2\sin2\phi,\qquad
\boxed{I_{2\omega}\propto\sin^22\phi,\quad\phi=45^\circ\pmod{90^\circ}.}
$$
The source <electric polarization> is therefore along $z$ and is largest along the in-plane diagonals. A radiated wave also needs a transverse component of this source: propagation along $z$ would not generate a propagating $z$-polarized output. Propagation in the $xy$ plane provides the simple strictly transverse example. For a tilted extraordinary wave, the electric displacement, rather than generally the entire <electric field>, is perpendicular to the <wave vector>; the tensor source must be projected onto that wave's allowed <polarization>. This distinction matters when using the type-I geometry below.
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