I am facing an exercise about maxima and minima for the function $$f(x, y, z) = xye^x - xyz$$
So the gradient is
$$\nabla f(x, y, z) = (ye^x + xye^x - yz, xe^x - xz, -xy)$$
The solutions I found are the points
$$P = (0, 0, z)$$ $$Q = (0, y, 1)$$ $$K = (x, 0, e^x)$$
Now the Hessian matrix reads
$$H = \begin{pmatrix} 2ye^x + xye^x & e^x + xe^x - z & -y \\ e^x + xe^x - z & 0 & -x \\ -y & -x & 0 \end{pmatrix} $$
The problem is that when I evaluate the Hessian in the points I have found, in all the three cases I find one zero eigenvalue, which means I cannot say anything about the point.
How is this possible? Is there a way to say anything about those points?
Let's study each of the three families of critical points you found.
For a $P$-type point $P=(0,0,z)$, we have $f(P)=0$. We also have $$f(\Delta x, \Delta y, z+\Delta z)=\Delta x \Delta y (\mathrm{e}^{\Delta x} - z-\Delta z), $$ which shows that $f$ can take both positive and negative values in any neighborhood of such points. Hence all $P$-type points are saddles.
For a $Q$-type point $Q=(0,y,1)$, we have $f(Q)=0$ and $$f(\Delta x, y+\Delta y, 1+\Delta z) = \Delta x(y+\Delta y)(\mathrm{e}^{\Delta x} - 1-\Delta z). $$ This also shows that $f$ takes both positive and negative values in neighborhoods of $Q$-type points. They are therefore saddles as well.
Finally, for a $K$-type point, $K=(x,0,\mathrm{e}^x)$, we have $f(K)=0$ and $$f(x+\Delta x, \Delta y, \mathrm{e}^x +\Delta z)=(x+\Delta x) \Delta y(\mathrm{e}^{x+\Delta x}-\mathrm{e}^{x}-\Delta z), $$
also giving saddles.