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The divergence is $y^2+x^2+z^2=r^2$. In spherical coordinates, the angular integral is $4\pi$ and the volume element contributes $r^{2}dr$. Thus

On $x\ge 1$, an integrating factor is $1/x$. Multiplication gives $(y/x{)}^{\prime }=1$, so the general solution is $y=x^2+Cx$. The initial value $y(1)=1$ forces $C=0$. Therefore

Taking each partial derivative of the scalar potential givesThe factor $5y^4$ in the middle component comes from the chain rule. The potential is smooth, so the curl of a gradient identity gives $\boxed{\nabla \times \mathbf{F}=0}$.

The characteristic polynomial is $(1-\lambda {)}^2-4=(\lambda -3)(\lambda +1)$. For $\lambda =3$ the components are equal; for $\lambda =-1$ they are opposite. After normalization, the eigenvalues and associated eigenvectors areAn overall sign change of either normalized eigenvector is equivalent.

The Taylor series of the logarithm givesThe first nonzero term is therefore $\boxed{x^2}$. It is positive on either side of the origin, so the stationary point is a strict local minimum. In fact $\log (1+x^2)\ge 0$, with equality only at zero, making it the global minimum as well.

Write $-8=8e^{i(\pi +2\pi k)}$. Its three roots have modulus $2$ and arguments $(\pi +2\pi k)/3$, with $k=0,1,2$. Thus all three complex roots areFurther integer values of $k$ repeat these roots.