Critical points of ${f}(x,y)=(x+2y)e^{x^{-2}-y^{-2}}$

Question

I know how to find the critical point but i can't solve the system. This is the function ${f}(x,y)=(x+2y)e^{x^{-2}-y^{-2}}$ This is the first derivative of x: $${f}'x = (x+2y)'e^{x^{-2}-y^{-2}} + (x+2y){e^{x^{-2}-y^{-2}}}'$$ $${f}'x = 1e^{x^{-2}-y^{-2}} + (x+2y)(e^{x^{-2}-y^{-2}})(-2x) $$ $${f}'x = 1e^{x^{-2}-y^{-2}} - (2x^2 + 4xy)(e^{x^{-2}-y^{-2}}$$ $${f}'x = e^{x^{-2}-y^{-2}} (1-2x^2 + 4xy)$$ This is the first derivative of y: $${f}'y = (x+2y)'e^{x^{-2}-y^{-2}} + (x+2y){e^{x^{-2}-y^{-2}}}'$$ $${f}'y = 2e^{x^{-2}-y^{-2}} + (x+2y)(e^{x^{-2}-y^{-2}})(-2y)$$ $${f}'y = 1e^{x^{-2}-y^{-2}} - (2xy + 4y^2)(e^{x^{-2}-y^{-2}}$$ $${f}'y = e^{x^{-2}-y^{-2}} (2-2xy - 4y)$$ The system is: \begin{vmatrix} {f}'x = e^{x^{-2}-y^{-2}} (1-2x^2 + 4xy) = 0\\ {f}'y =e^{x^{-2}-y^{-2}} (2-2xy - 4y)= 0 \end{vmatrix} The problem is that I can't solve the system...