\(f(x,y)=9e^y(y^2-x^2)\)
\(abla f=\left\langle-18xe^y,9ye^y(2+y)\right\rangle\)
Critical points occur where the gradient is zero. This is guaranteed whenever \(x=0\) and either \(y=0\) or \(y=-2\).
The Hessian matrix for this function looks like
\(H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-18e^y&-18xe^y\\-18xe^y&9e^y(2-x^2+4y+y^2)\end{bmatrix}\)
and has determinant
\(|H(x,y)|=-162e^{2y}(2+x^2+4y+y^2)\)
Maxima occur whenever the determinant is positive and \(f_{xx}<0\). Minima occur whenever both the determinant and \(f_{xx}\) are positive. Saddle points occur whenever the determinant is negative.
At \((0,0)\), you have a saddle point since the determinant reduces to -324, so \((0,0,0)\) is the saddle point.
At \((0,-2)\), the determinant is \(\dfrac{324}{e^4}>0\) and \(f_{xx}(0,-2)=-\dfrac{18}{e^2}<0\), so \(\left(0,-2,\dfrac{36}{e^2}\right)\) is a local maximum.
No other critical points remain, so you're done.
