Consider the function $f(x,y) = x^3 + e^{3y}-3xe^y$. Show that $f$ has exactly one critical point and that this point is a local minimizer, but not a global minimizer.
I have attempted this, but it seems that I have made a mistake somewhere along the way. Can someone please help me find where I went wrong?
My attempt:
\begin{align}\nabla f = \begin{bmatrix}3x^2 -3e^y \\3e^{3y} -3xe^y\end{bmatrix} = \bar{0}\end{align} Thus giving the following two homogeneous equations \begin{align}3x^2 - 3e^y &= 0\\ 3e^{3y} - 3xe^y &=0\end{align}
From the second equation we find that $$x = e^{2y}$$ and upon substitution into the first equation we find that $$e^y(e^{3y} -1) =0$$
Notice however that $e^y \neq 0$ and we must that have that $e^{3y}=1 \implies y=0$ and thus $x =1$
Our critical point is thus $(x,y) = (1,0)$.
How consider the Hessian
\begin{align}H_f &= \begin{bmatrix}6x & -3e^y \\ -3e^y & 9e^{3y}-3xe^y\end{bmatrix}\end{align} Evaluated at our critical point we obtain $$H = \begin{bmatrix}6 & -3 \\ -3& 6\end{bmatrix}$$
We can then deduce that $H_f$ is positive definite and thus our critical point must be a strict global minimizer. This, however is the exact opposite of what the question stated?
Edit: This is the theorem in my notes that I tried to use:
Your conclusion is wrong. If the Hessian is positive definite, the critical point is a local minimizer.