I have a function $$f: \mathbb{R^3 \rightarrow R}, f(x,y,z) = (x+xy+yz)e^x$$ Through the gradient I have found the stationary point $(-1,0,1)$. I want to conclude what kind of critical point this is through the second partial derivative test.
I have the quadratic form: $$Q(h,k,l)=f_{xx}h^2+f_{yy}k^2+f_{zz}l^2+2(f_{xy}hk+f_{xz}hl)+f_{yz}kl$$ Plugging in the values at $(-1,0,1)$ I get: $$f_{xx}=f_{xy}=f_{yz}=\frac{1}{e}, f_{yy}=f_{zz}=f_{xz}=0$$
Which $\implies$ $$Q(h,k,l)=\frac{1}{e}(h^2+2(hk+kl))$$
But it seems that for different values of $h,k$ and $l$, I can get positive, negative and zero for $Q$. I am not sure how to interpret this result.
Edit:________________________________________________________________________
After a suggestion of Hessian method, I instead tried to solve it by seeing looking for the sign of: $$\begin{bmatrix}x+1&y&z-1\end{bmatrix} \begin{bmatrix} \frac{1}{e} & \frac{1}{e}&0\\\frac{1}{e}&0&\frac{1}{e}\\0&\frac{1}{e}&0\end{bmatrix}\begin{bmatrix}x+1\\y\\z-1\end{bmatrix}$$ Which equals: $$\frac{1}{e}((x+y+1)(x+1)+y(x+z)+y(z-1))$$
Finally I conclude that the matrix is indefinite, since it can attain both negative and positive values and $(-1,0,1)$ is therefore a saddle point.
Please let me know if there is an error in the reasoning.