According to Wolfram, the following matrix has complex eigenvalues.
Symmetric matrices have real eigenvalues, so I’m not sure what I’m failing to understand. The matrix is the shape operator of the hypersurface $f(x,y,z) = (x,y,z,xyz)$. I don’t see why this manifold would have complex-valued principal curvatures.
The comment exchange already revealed that the appearance of complex numbers in the formulas, the third root of unity in particular, was an artefact produced by Cardano's formula. A trigonometric way of solving a cubic equation makes it abundantly clear that the eigenvalues of this matrix are all real.
Let $A$ be the simpler matrix $$ A=\left( \begin{array}{ccc} 0 & z^2 & y^2 \\ z^2 & 0 & x^2 \\ y^2 & x^2 & 0 \\ \end{array} \right).$$ Your matrix is a scalar multiple of $A$, so it suffices to find the eigenvalues of $A$. The first thing to do is to calculate the characteristic polynomial $$ \chi(T)=\det(T I_3-A)=T^3-(x^4+y^4+z^4)T-2x^2y^2z^2.\tag{1} $$ The trigonometric method of solving a (depressed) cubic relies on the identity $$ 4\cos^3x-3\cos x=\cos 3x.\tag{2} $$ The way to use it is to find a suitable new variable $u$ such that the original equation becomes $$ 4u^3-3u=v,\tag{3} $$ where hopefully $v$ has absolute value $\le1$. Whenever that is the case, we can write $v=\cos\beta$ for some real angle $\beta$, and $(2)$ then implies that the solutions of $(3)$ are $u_k=\cos\frac{\beta+k\cdot2\pi}3$, $k=0,1,2.$
A comparison of $(1)$ and $(3)$ shows that we are looking for a substitution $\alpha u=T$ such that $$ \frac{\alpha^2}{x^4+y^4+z^4}=\frac43. $$ In other words, we want $$\alpha=\frac2{\sqrt3}\sqrt{x^4+y^4+z^4}.\tag{4}$$ Plugging in $T=u/\alpha$ with the choice $(4)$ of $\alpha$ does yield after simple cleaning up the equation $$ 4u^3-3u=\frac{3\sqrt3 x^2y^2z^2}{(x^4+y^4+z^4)^{3/2}}.\tag{5} $$
According to our plan we next need to verify that the right hand side of $(5)$ has absolute value at most $1$. Thankfully this is straight forward. By the AM-GM -inequality we have $$ x^4+y^4+z^4\ge3(x^4y^4z^4)^{1/3}. $$ Everything in sight is non-negative, so we can raise this to power $3/2$, and arrive at the inequality, valid for all real $x,y,z$ $$ (x^4+y^4+z^4)^{3/2}\ge 3\sqrt3x^2y^2z^2. $$ This proves that the right hand side of $(5)$ is in the interval $[0,1]$, and hence the desired angle $\beta$ such that $\cos\beta=3\sqrt3x^2y^2z^2/(x^4+y^4+z^4)^{3/2}$ exists.
The eigenvalues $\lambda_k, k=0,1,2$, of $A$ are thus $$ \lambda_k=\frac2{\sqrt3}\sqrt{x^4+y^4+z^4}\cos\frac{\beta+k\cdot2\pi}3. $$ The eigenvalues $\Lambda_k=\gamma \lambda_k$ of the original matrix are gotten from these with the scaling factor $\gamma=xyz/(x^2y^2+y^2z^2+z^2x^2+1)^{3/2}.$
These are all manifestly real.