Answer
So, $f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=1$ and $y=1$.
Work Step by Step
Notice that $f(x,y)$ is well defined for $x\neq0$ and $y\neq0.$
First we will find both partial derivatives. The partial derivative with respect to x is:
$$f_x(x,y)=\frac{\partial}{\partial x}\Big(\frac{1}{x}+\frac{1}{y}+xy\Big)=
-\frac{1}{x^2}+y$$
The partial derivative with respect to y is:
$$f_y(x,y)=\frac{\partial}{\partial y}\Big(\frac{1}{x}+\frac{1}{y}+xy\Big)=-\frac{1}{y^2}+x$$
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 $0$:
$$f_x(x,y)=y-\frac{1}{x^2}=0,f_y(x,y)=x-\frac{1}{y^2}=0$$
From the first equation we have: $y=\frac{1}{x^2}$ . Putting this into the second equation we get:
$$x-\frac{1}{(\frac{1}{x^2})^2}=0\Rightarrow x-x^4=0\Rightarrow x(1-x^3)=0\Rightarrow x=0 \vee1-x^3=0\Rightarrow x=0\vee x^3=1\Rightarrow x=0 \vee x=1$$.
Since for $x=0$ function $f$ is not defined, we conclude that it must be $x=1$. Now we have that $y=\frac{1}{x^2}=\frac{1}{1^2}=1$
So, $f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=1$ and $y=1$.
