Calculating Stationary Points of $f(x,y)=e^{-(x^2+y^2)}$

Question

Here's my attempt so far:

$$f_x(x,y)=-2xe^{-(x^2+y^2)}$$ $$f_y(x,y)=-2ye^{-(x^2+y^2)}$$ I tried equating both the partial derivatives to $0$, and the only solution I seem to get is $(x,y)=(0,0)$. Are there any more solutions to this?

For more information, here are the second partial derivatives: $$f_{xx}(x,y)=e^{-(x^2+y^2)}(-2+4x^2)$$ $$f_{yy}(x,y)=e^{-(x^2+y^2)}(-2+4y^2)$$ $$f_{xy}(x,y)=-2ye^{-(x^2+y^2)}(-2+4x^2)$$

Answer 1

Summarising the discussion in the comments, there is just a single stationary point of $f$ at $f(x,y)=(0,0)$, and that is a global maximum of the function.