Answer
$f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=0$ and $y=0$.
Work Step by Step
First we will find both partial derivatives. The partial derivative with respect to x is:
$$f_x(x,y)=\frac{\partial }{\partial x}(e^{x^2+xy+y^2})=e^{x^2+xy+y^2}\frac{\partial }{\partial x}(x^2+xy+y^2)=e^{x^2+xy+y^2}(2x+y)$$
The partial derivative with respect to y is:
$$f_y(x,y)=\frac{\partial }{\partial y}(e^{x^2+xy+y^2})=e^{x^2+xy+y^2}\frac{\partial }{\partial y}(x^2+xy+y^2)=e^{x^2+xy+y^2}(x+2y)$$
To find all values of $x$ and $y$ such that $f_x(x,y)=0$ and $f_y(x,y)=0$ we will equate both partial derivatives with $0$:
$$f_x(x,y)=e^{x^2+xy+y^2}(2x+y)=0,f_y(x,y)=e^{x^2+xy+y^2}(x+2y)=0$$
Because exponential function is never equal to $0$ we can cancel it, so our equations become:
$$2x+y=0,x+2y=0$$
From the first equation we have: $y=-2x$. Putting this into the second equation we get:
$$x+2\cdot(-2x)=0\Rightarrow-3x=0\Rightarrow x=0,$$
which gives us $y=-2x=-2\cdot0=0.$
So, $f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=0$ and $y=0$.
