Finding critical points of $e^y(y^2-x^2)$

Question

I'm going back doing extra problems in areas I had trouble with and am doing a problem asking to find and describe the critical points of $f(x,y)=e^y(y^2-x^2)$

I have

$f_x=-2x,$ and $f_y=e^yy^2+2ye^y$ by the product rule.

$f_{xx}=-2$, $f_{yy}=e^y(4y+y^2+2)$, $f_{xy}=0$

and setting $f_x$ and $f_y$ to zero, found $x=0, y=1||2$

plugging each into $f_{xx}f_{yy}-f_{xy}^2$ I got $-4$ for $(0,0)$, negative and thus a saddle point. For $(0,2)$ I got $-28e^2$ which would be negative, so another saddle point. Since in the function x and y are squared I did $-2$ as well for y, getting $(0,-2)$ as another saddle point.

Can someone tell me if I'm doing this correctly?

Answer 1

I think that you made a mistake with the first partial derivative and it propagated through all the next steps. $$f=e^y \left(y^2-x^2\right)$$ $$f_x=-2 x e^y\qquad f_y=e^y \left(y^2+2 y-x^2\right) $$ $$f_{xx}=-2 e^y\qquad f_{yy}=e^y(y^2+4 y-x^2+2)\qquad f_{xy}=-2 x e^y$$