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:
$$f_x(x,y)=\frac{\partial}{\partial x}(\ln(x^2+y^2+1))=\frac{1}{x^2+y^2+1}\frac{\partial}{\partial x}(x^2+y^2+1)=\frac{2x}{x^2+y^2+1}$$
The partial derivative with respect to y is:
$$f_y(x,y)=\frac{\partial}{\partial y}(\ln(x^2+y^2+1))=\frac{1}{x^2+y^2+1}\frac{\partial}{\partial y}(x^2+y^2+1)=\frac{2y}{x^2+y^2+1}$$
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 $0$:
$$f_x(x,y)=\frac{2x}{x^2+y^2+1}=0,f_x(x,y)=\frac{2y}{x^2+y^2+1}=0$$
From the first equation we have: $2x=0\Rightarrow x=0$ and from the second one we have $2y=0\Rightarrow y=0$
So, $f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=0$ and $y=0$.
