I have that
\begin{array}{lcl} f'_{x} & = & (1+x+y+xy+yz)e^x \\ f'_{y} & = & (x+z)e^x \\ f'_{z} & = & ye^x \end{array}
Solving all these partials, simultaneously equal to zero I get that there exists only one statinary point that is $(x,y,z)=(-1,0,1).$ Now I want to determine the nature of this point by examining the quadratic form $Q(h,k,l).$ For the second derivatives I have
\begin{array}{lcl} f''_{xx} & = & (2+2y+x+xy+yz)e^x \\ f''_{yy} & = & 0 \\ f''_{zz} & = & 0 \\ f''_{xy} & = & (1+x+z)e^x \\ f''_{yz} & = & e^x \\ f''_{xz} & = & ye^x \\ \end{array}
Which gives
\begin{array}{lcl} Q(h,k,l) & = & f''_{xx}(-1,0,1)h^2+f''_{yy}(-1,0,1)k^2+f''_{zz}(-1,0,1)l^2 +\\ & + & 2f''_{xy}(-1,0,1)hk +2f''_{yz}(-1,0,1)hl+2f''_{xz}(-1,0,1)kl \\ & = & \frac{1}{e}h^2+\frac{2}{e}hk+\frac{2}{e}hl=\frac{1}{e}(h^2+2hk+2hl). \end{array}
Using the trinomial squared rule I can rewrite the last equality as
$$\frac{1}{e}\left[(h+k+l)^2-k^2-l^2-2kl\right]=\frac{1}{e}\left[(h+k+l)^2-(k+l)^2\right],$$
which means that this quadratic form is positively indefnite since it can attain both negative and positive values which in turn implies that this is a saddle point.
Question: Am I doing this right?