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Let $f(x, y) = 2xye^{-x^2-y^2}.$ Find critical points. Use second derivative to to determine whether a critical point corresponds to local min, max or saddle.

Hi so for this question I am able to get my critical points by taking all the partial derivatives but when I get to inputting my critical points into the equation $f_{xx}f_{yy}-(f_{xy})^2$. I don't get the right answer. For instance one of the critical points is $(0, 0)$ but when I input it into the equation I get $0$ which is not correct as it is supposed to be $-4$. Can someone show me how to use the second derivative for this question.

EDIT: I took the partial derivatives to get Fxx= 4(2x^2-3)xye^(-x^2-y^2) Fyy = 4(2y^2-3)xye^(-x^2-y^2) Fxy = 2(2y^2-1)xe^(-x^2-y^2)

D(x, y) = fxx*fyy - (fxy)^2

Therefore, whenever I put the required equations in and set them to 0 I do not get the answer expected which is -4.

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    $\begingroup$ Please edit your post to include your work. Otherwise we have to be mind readers to figure out why you don't get the right answer. $\endgroup$ – Ted Shifrin Jan 20 '20 at 3:11
  • $\begingroup$ The question has been EDITED I included the partial derivatives and the formula used to check what D(x, y) is equal to. $\endgroup$ – OGK Jan 20 '20 at 6:02
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Your value for $f_{xy}$ cannot be correct, as $x$ and $y$ do not appear symmetrically in it. (Note that $f(x,y)=f(y,x)$, so everything should be symmetric.) Indeed, $$f_{xy} = 2(2x^2-1)(2y^2-1)e^{-(x^2+y^2)},$$ so $f_{xy}(0,0) = 2$.

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