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Put $h=x-\pi /4$. The trigonometric identities give $\cos x\sin x=\frac{1}{2}\sin 2x=\frac{1}{2}\cos 2h$. Expanding cosine at zero givesThe linear and cubic terms vanish, so these are the first two nonzero Taylor series terms.

The characteristic polynomial is $(1-\lambda )(4-\lambda )-4=\lambda (\lambda -5)$. Solving each eigenvalue equation gives the eigenvalues and eigenvectorsFor unit-length representatives divide each displayed direction by $\sqrt{5}$. The two eigendirections are orthogonal, consistently with the matrix being a symmetric matrix.

The gradient is $\nabla \varphi =(2xy+z,x^2,x)$, with value $(3,1,1)$ at the point. Normalize the direction to a unit vector: $\hat{\mathbf{s}}=(2,0,1)/\sqrt{5}$. The directional derivative is therefore

Multiply numerator and denominator by the complex conjugate of the denominator:Thus the real part and imaginary part are $\boxed{\mathrm{Re}=-5/13,\mathrm{Im}=-12/13}$.